Chapter 1: The Genetic Approach to Biology
Questions for Chapter 1 from text: 2, 4, 8, 11, 17
Gene - basic elements of the system of inherited information; the fundamental physical and functional unit
of heredity, which carries information from one generation to the next; a segment of DNA composed of a
transcribed region and a regulatory sequence that makes transcription possible
Allele - one of the different forms of a gene that can exist at a single locus
Phenotype - the form taken by some character or group of characters in a specific individual; the
detectable outward manifestations of a specific genotype
Genotype - the specific allelic composition of a cell, either of the entire cell or, more commonly, of a certain
gene or a set of genes
Mutation - the process that produces a gene or a chromosome set differing from that of the wild type; the
gene or chromosome set that results from such a process
Organisms
Give rise to like organisms (reproduce)
Respond to inputs:
- In unicellular organisms, responses to different inputs possible Unicellular responds to environment
by changes within the cell
Eg) high temp = denatured proteins
Eg) when low food = optimizes ability to absorb and retain nutrients
- In multicellular organisms, cells derived from a single cell develop complex and diverse structures
Multicellular organisms' cells are derived from a single cell develop complex and diverse structures
Evolve over time
Inherited units = genes
Idea of inheritance by Gregor Mendel
DNA is the molecule which makes a gene
Idea of genes arose in early 1900s
What key properties must hereditary material possess?
1. Ability to replicate faithfully (inheritance)
2. Provide an extraordinary diversity of information that can be
3. translated (changed) into structure and function of cells
4. Ability to change over time (evolution)
Mutation combined with natural selection allows change over time
Humans and chimps differ in 1% of DNA
The genomes of humans and chimpanzees differ by only a small percentage of nucleotides
1. DNA Replication
Each DNA strand is the template for the production of a new strand - result is two DNA molecules identical
to the original
DNA molecule is a double-stranded helix
Nucleotides are joined by weak hydrogen bonds that can be separated by DNA polymerase or helicase
Nucleotides (ATGC) are joined to strand at sulfur and phosphorus
Strong covalent bonds occur along each strand (sulfur and phosporus)
2. Diversity of Genes
There are four kinds of nucleotide (ATGC) within a molecule
Diversity is possible because each gene is made up of thousands of nucleotides
3. Translation of information
DNA > RNA > protein
Transcription of DNA from a gene makes RNA
Eg) Ribosomal RNA does work within cell as RNA
Eg) Messanger RNA helps make proteins
4. Mutation
Mutation combined with natural selection allows change over time
- Errors during replication
(enzymes recognize simple errors of Purine-Pyramidine pairing)
- Mutation due to exposure to mutagens
(chemical, x-ray, UV)
Hereditary material (DNA) possesses 3 key properties:
1. DNA replication allows faithful replication that enables inheritance of information both between and
within generations
2. Variations in DNA sequence provide an extraordinary diversity of information that can be translated
via RNA and protein into cellular structure and function
3. Mutatation of DNA allows change over time
Study of genes involves studying the effects of mutant forms of a gene (alleles) on the organism
(Physiology, Biochem, development, disease, behaviour, ecology, evolution, etc.)
Studying genetic diversity leads geneticists to understand biological processes
genetic diversity (different genotypes) results in observable differences in organisms' morphology,
physiology, development, ability to withstand environmental stress (phenotype)
Albino
-During the process genotype > transcription > translation > cellular phenotype result something goes
wrong.
-A mutant allele (genotype) produces a malfunctioning enzyme that leads to albinism (phenotype) (Figure
1-7)
-A malfunctioning enzyme leads to albinism
genetic diversity:
alleles are different forms of a gene. An allelle is a form/ variant of a gene
genotype:
Normal DNA sequence = wild type allele
Altered DNA sequence = mutant allele
Phenotype:
Normal morphology = wild type phenotype
Abnormal morphology = mutant phenotype
Model organisms are commonly used in genetic analysis
Need enzymes to cut paste and arrange vectors of DNA
Need ways of manipulating and testing organisms
Tend to use same species of organism such as lab rats, small plants
These model organisms share the following characteristics:
Short generation time
Small size
Easily maintained
Large numbers of progeny
Small genome
Representative (central to taxonomic group)
Model Organisms in Genetics:
Virus: bacteriophage lambda and its bacterial host Escherichia coli
Fungus: Neurospora crassa
Budding Yeast: Saccharomyces cerevisiae
Plants: Zea maize; Arabidopsis thaliana
Insect: Drosophila melanogaster
Nematodes: Caenorhabditis elegans (completely transparent under microscope makes it easy to
follow cell development)
Vertebrates: Mus musculus (mouse), Xenopus laevis (frog)
Clicker
If a DNA molecule is 100 bases long, how many different combinations are possible?
4^100
33 amino acids are made from 100 bases
Example: Consistency
Let A = yellow and a= green; also, let the gene symbol be gene A
We could also use: P, Y, X, Z
Alleles: yel, YEL, yel-1, yel+, yela
It doesn't matter what symbol you use, just be consistent
Test Cross:
When an individual of unknown genotype is crossed with a homozygous recessive individual to determine
unknown's genotype through phenotypic result. The test cross progeny are in equal frequency, consistent
with Mendels' hypothesis:
Suggest parents contain 2 alleles, gametes contain
Reproduction (meiosis specifically) accounts for this hypothesis because of alternating mitotic and meiotic
divisions during eukaryotic life cycles
Diploid > meiosis > haploid > fusion > diploid > mitosis
Chapter 2: Single-Gene Inheritance
The nuclear genome: all chromosomes of nuclear DNA, genes, and intergenic regions
(not mitochondrial, chloroplast or plasmal DNA/RNA for nuclear genome)
"-some" means nucleic acid plus proteins
centromeres important in movement/ segregation of chromosomes
Single-celled organism have smaller genomes, but also pack genes closer together, smaller intergenic
regions
Looking at a piece of DNA being transcribed:
Introns taken out and remaining pieces spliced together
DNA
Start
Exon
Stop
intron
exon
intron
pre-mRNA
Chromosomal DNA wrapped around histones
8 histones come together in a protein group called "octamer"
These 8 polypeptides are for different types of histones: 2x(H2A, H213, H3, H4)
Nucleosome: octamer wound around two times
Chromatin: DNA and associated proteins
Chromosomal DNA is wrapped around histone octamer (nucleosome), nucleosomes associate with and
coil
around H1 histone
The nucleosome is 10 nm in diameter
Histone octamers coil until 30 nm in diameter to made a solenoid
Attachment to scaffold results in supercoiling
Scaffold is a set of proteins
eg) topoisomerase
Telomere = chromosome end
Centromere = constricted area of chromosome
Nucleolar organizer - tandem repeats of genes encoding rRNA
Nucleolus - RNA and protein associated; located around nucleolar organizer
Heterochromatin - dense chromatin
Euchromatin - less dense chromatin
Diploid genome - 2 sets of chromosomes
Eg) if 2n=6 there are 3 homologous pairs
Gregor Mendel - founder/ father of genetics
Firm understanding in physics and mathematics
Garden pea:
model organism
Diploid
fast generation time
readily available, with multiple varieties
easily grown
Seedless watermelon is polyploid and infertile
Wheat is a hexaploid result of three wild species coming together
Three sets of chromosomes doubled in order to multiply
Traits can be selected for such as gluten protein, stronger shells around seeds
Bigger plants and bigger yield
In peas, both cross=pollination and self-polination (selfing) are possible
Mendel studied seven phenotypic pairs that were controlled by single genes:
Round or wrinkled seed
Yellow or green seeds
Purple or white flowers
Inflated or pinched ripe pods
Green or yellow pods
Axial or terminal flowers
Long or short stem
Mendel generated pure breeding lines
If after 7 generations of selfing all populations were pure
P
Pure Yellow
YY
Pure Green
yy
F1
all yellow offspring: dominant phenotype Yy
F2
Selfing of F1
3 dominant yellow : 1 recessive green
YY or Yy
yy
Mendel's explanation - each parent contains 2 copies of each factor (each parent is diploid with 2 alleles of
each gene)
Parental lines have 2 identical alleles (homozygous)
Each parent has 2 copies hereditary units
2 alleles of the gene
Each parent contributes one unit to the offspring (diploid>haploid reproduction)
Meiosis produces one haploid gamete from each parent
YY - homozygous dominant
Yy - heterozygous
yy - homozygous recessive
The F1 of Yy has 2 different alleles (heterozygous: monohybrid)
Monohybrid: one gene involved for character
Dominant allele - phenotype is expressed in a heterozygote
Recessive allele - phenotype is not expressed in a heterozygote
Male Y
Yellow
Parent y
RESULT:
Yellow Female Parent
Y
y
YY
Yy
Yy
3: 1 Ratio of yellow to green
yy
3x[½ Y ½ Y = ¼ Y]
Yellow
¾ dominant
1x[ ½ y ½ y = ¼ y ] green
¼ recessive
Frequency of 4 combinations is product of frequencies of each gamete frequency (product rule): two
independently occurring events
Independent assortment = separation
CHAPTER 2
Definitions:
genome - the complement of genetic material in the chromosomes
Gene - the fundamental unit of heredity (in molecular terms, a transcribed segment of DNA and the region
which enables transcription)
Intergenic region - region of DNA between genes
Intron - (intervening sequence) a segment of the gene that is initially transcribed, but is not in the final
mRNA product
Exon - a segment of the gene that is translated into the protein
nucleosome - 8 histone polypeptides (octamer) wrapped in DNA
Telomere - end of chromosome
centromere - part of chromosome to which the spindle fibres attach via the kinetochore
heterochromatin - densely staining, highly condensed regions of the chromosome that are usually not
transcribed
Euchromatin - a less condensed region, thought to contain genes that can be transcribed
haploid - a cell or individual having one chromosome set
diploid - a cell or individual having two chromosome sets
homozygous (= true-breeding): an individual having identical alleles of a gene
heterozygous: an individual having different alleles of a gene
monohybrid: an individual heterozygous at one gene
first filial (F1) generation - the first generation resulting from a controlled cross between two known
parents (P)
second filial (F1) generation - the second generation resulting from a controlled cross between two
known parents (P)
test cross - a cross between an individual of unknown genotype and an individual homozygous recessive
for a particular gene(s)
product rule: the probability that two independent events occur together is equal to the product of their
individual probabilities
sum rule: the probability that either of two mutually exclusive events occurs is equal to the sum of their
individual probabilities
genetic symbol: a symbol designating a gene which usually relates to the characteristic controlled by that
gene (e.g. Y = yellow gene)
allelic symbol: a symbol designating an allele which is a derivative of the gene symbol (e.g. Y = yellow
allele, y = green allele)
sex chromosome - a chromosome whose presence is associated with a particular sex
autosome - a chromosome that is not a sex chromosome
sex linkage - the location of a gene on a sex chromosome
hemizygous - a gene present in only one copy in a diploid organism
Text from Lectures:
Single-Gene Inheritance
The nuclear genome (Figure 2-2)
DNA is condensed to form chromosomes
Representative chromosomal landscapes in different organisms;
note differences in gene density, presence of introns (Figure 2-7)
2 regions of human chromosome 21 showing banding patterns, gees with known transcription
products (arrows) Figure 2-8
A diploid genome (Indian mutjac, 2n=6) visualized by dye specific to each chromosome (Figure 2-3)
Interphase cell, chromosomes are in particular domains of the nucleus
Cell beginning division phase, chromosomes are compact, clearly showing 2 members of each
homologous pair
Chromosomal DNA is wrapped around histones
Chromatin = DNA + associated proteins (Figure 2-4a)
Space-filling model shows DNA wrapped around histone octomer to form nucleosomes
Histone octamer (green and blue) = 2x H2A, H2B, H3, H4
Yellow = histone H1
Chromosomal DNA is wrapped around histone octamer (nucleosome), nucleosomes associate with and
coil around H1 histones (Figure 2-4b)
6 nuclesomes + 6 H1 histones = 1 coil
note compaction
Attachment to scaffold results in supercoiling (Figure 2-5)
Some landmarks of tomato chromosome 2 (Figure 2-6)
Telomere = chromosome end
centromere = constricted area of chromosome
Nucleolar organizer - tandem repeats of genes encoding rRNA
Heterochromatin - dense chromatin
euchromatin - less dense chromatin
The founder of genetics, Gregor Mendel Figure 1-1
Mendel choose garden pea as a "model organism"
Diploid
Fast generation time
Readily available, with multiple varieties
Easily grown
In peas, both cross-pollination and self-pollination (selfing) are possible (Figure 2-10)
Mendel studied seven phenotypic pairs (Figure 2-9) with clearly contrasting and easily distinguished
phenotypes
Grew plants for several successive generations to be certain lines were true-breeding
For each characteristic, Mendel crossed pairs of pure-breeding parents Figure 2-12 part 1
For all crosses the first filial generation (F1), showed only one phenotype; defined as the dominant
phenotype, the other phenotype defined as recessive
(Figure 2-12 part 2)
For all crosses, the second filial generation (F2), showed only two phenotypes, and the ratio was
always 3:1 (the larger class was the same phenotype as that in the F1; the dominant phenotype) (Figure 212 part 3)
Mendel's explanation - each parent contains 2 copies of each factor; each parent is diploid with 2
alleles of each gene)
Parental lines have two identical alleles, they are homozygous (Figure 2-12 part 5)
Mendel's explanation - each parent contributes 1 copy of each factor to offspring with equal
frequency (Mendel's law of equal segregation); gametes are haploid, fuse to form diploid offspring
The F1 has 2 different alleles (heterozygous; monohybrid)
Dominant allele - phenotype is expressed in a heterozygote
Recessive allele - phenotype is not expressed in a heterozygote
Figure 2-12 part 6
Mendel's explanation - F1 contributes 1 allele to F2 offspring - alleles fuse randomly with other
(haploid gametes fuse to generate diploid offspring)
Figure 2-12 part 7
Mendel's explanation - random fusion of gametes results in 4 combinations:
Figure 2-12 part 11
Frequency of 4 combinations is product of frequencies of each gamete frequency (product rule):
i.e. ½ x ½ = ¼
Yields a genotypic frequency of 1YY : 2Yy : 1 yy
Yields a phenotypic frequency of 3 Y_ : 1 yy
Figure 2-12 part 12
Next, Mendel performed a test cross - crossing one individual to an individual that is homozygous
recessive Figure 2-12 part 4
If the F1 plant is heterozygous, it should generate two types of gametes with equal frequency
I.e. F1 gametes are Y = ½, y= ½
Tester parent (y) generates only one type of gamete, y
Therefore, phenotype of test cross progeny should reflect frequency of gametes from the F1
i.e. 1 Yy : 1 yy
(Figure 2-12 part 13)
Test cross progeny are in equal frequency, consistent with Mendel's hypothesis
(Figure 2-12 part 15)
Mendel's crosses are experiments that allow determination of number of genes, relationships
amongst alleles, and genotypes of individuals
(Figure 2-11)
(note, the progeny of the left cross in Figure 2-11 is not the F2, rather it is the test cross progeny)
Clicker Question #1:
in cross 6, what is the dominant phenotype?
a) axial
b) terminal
Clicker question #2:
Two different alleles of the gene for plant height exist in peas. True breeding individuals may be either tall
or short. Two tall individuals are crossed to one another, and the progeny are ¾ tall and ¼ short. The best
explanation for this result is:
a) both tall individuals are homozygous, the tall allele is dominant
b) both tall individuals are homozygous, the short allele is dominant.
c) both tall individuals are heterozygous, the short allele is dominant.
d) both tall individuals are heterozygous, the tall allele is dominant.
E) none of the above accurately explains the results
Clicker Question #3
If a test cross is done to a heterozygous individual with yellow seeds, what phenotype(s) will the progeny
be, and in what ratio?
A) All yellow
B) All green
C) 3 yellow: 1 green
D) 1 yellow : 1 green
E) None of the above
Stages of Mitosis
Interphase
Early mitotic prophase
Late mitotic prophase
Mitotic metaphase
Mitotic anaphase
Mitotic telophase
Meiosis- two sequential nuclear divisions:
MI - segregates homologous chromosomes
MII - segregates chromatids
Generates haploid daughter cells
Stages of Meiosis
Prophase I
Chromosomes pair to form bivalent (2 dyads or double stranded = tetrad)
Leptotene, zygotene, pachytene, diplotene
Crossing over occurs
Homologous chromosomes pair (fundamental difference between mitosis and meiosis, as well as
between
meiosis I and meiosis II)
Homologous chromosomes: Same genes, possibly different alleles
Metaphase I
Paired homologous chromosomes (bivalent) align at metaphase plate
Anaphase I
Segregation
Telophase I
Reformation of nuclear membrane, cell division may occur
Cells are haploid from this point onwards
Prophase II
Chromosomes recontract
Metaphase II
Chromosomes (dyads) line up
Anaphase II
Centromeres split and sister chromatids separate
Telophase II
Nuclei reform in 4 haploid cells
Comparision of meiosis and mitosis
Mitosis - separation of chromatids
Meiosis I and II
I: separation of homologous chromosomes
II: separation of sister chromatids
If we consider the behaviour of genes on chromosomes, alleles segregate during Meiosis I (separation of
homologous chromosomes)
Mendels' ratios predict the segregation
Demonstration of equal segregation is more straightforward within a haploid life cycle such as in yeast
Two mating types fuse to form meiocyte
You cannot assess (score) the phenotype of the haploid generation
Assigned problems:
9, 17, 20, 21, 22, 26, 28, 32, 36
Chapter 2b - Mitosis and Meiosis
Definitions:
Mitosis - cell division that produces 2 daughter cells with identical nuclei to the parent cell
Meiosis - a process consisting of 2 consecutive cell divisions that produces daughter cells (gametes or
sexual spores) with half the genetic material of the parent
Meiocyte - a diploid cell about to undergo meiosis
Chromatid - one of 2 daughter DNA molecules produced by chromosome replication
Dyad - a chromosome with 2 chromatids; a double-stranded chromosomes
Homologous pair - chromosomes that pair with each other at meiosis and contain the same genes (may
have different alleles)
Bivalent - two homologous chromosomes paired in Meiosis I
Synaptonemal complex - complex that allows chromosome pairing during Prophase I
Kinetochore - proteins that attach the centromere to the kinetochore microtubules
Reduction division - division that results in cells with one member of each homologous pair (i.e. Meiosis
I)
Crossing over - the exchange between homologous chromosomes of parts of a chromatid through
breakage and rejoining
Mendel's hypotheses:
suggest parents contain 2 alleles, gametes 1 allele
gametes fuse to regenerate diploid in progeny
what processes explain these ideas?
what mechanism accounts for separation of alleles into gametes?
Alternating mitotic and meiotic divisions during eukaryotic life cycles account for Mendel's diploid
to haploid transitions (Figure 2-14)
Stages of the asexual cell cycle (figure 2-13)
During the S phase, DNA molecules replicate to form identical chromatids (Figure 2-18)
Replicated chromosomes are segregated into daughter cells in an organized fashion during
Mitosis - generates cells identical to mother cell
Meiosis - generates haploid cells from diploid
Stages of Mitosis (Box 2-1)
Interphase = G1, S, G2 (growth + DNA replication)
Figure 2-15
Prophase = chromosome condensation, break down of nuclear membrane (Figure 2-15)
Metaphase = alignment of chromsomes along metaphase plate (Figure 2-15)
Alignment requires action of kinetochore microtubules, attached to centromeres, (Figure 2-23)
Anaphase = separation of chromosomes (Figure 2-15)
Telophase = cytokinesis, reformation of nuclear membrane (Figure 2-15)
Mitosis generates daughter cells of identical ploidy as original cell (Figure 2-15)
Meiosis - two sequential nuclear divisions:
MI - segregates homologous chromosomes
MII - segregates chromatids
Generates haploid daughter cells (Figure 2-15)
Stages of Meiosis (Box 2-2)
Prophase (Figure 2-15 part 7):
Chromosomes pair to form bivalent (2 dyads = tetrad)
Crossing over occurs
Synaptonemal complexes at meiosis (Figure 2-16a)
Figure 2-16b
MetaphaseI - paired homologous chromosomes (bivalents) align at metaphase plate (Figure 2-15)
Anaphase I - segregation of homologous chromosomes (dyads) due to kinetochore shortening (Figure 215)
Telophase I - reformation of nuclear membrane, cell division may occur - cells are haploid from this point
onward (Figure 2-15)
Prophase II - chromosomes recontract (Figure 2-15)
Metaphase II - chromosomes (dyads) align at the metaphase plate (Figure 2-15)
Anaphase II - centromeres split and sister chromatids separate (Figure 2-15)
Telophase II - nuclei reform in 4 haploid cells (Figure 2-15)
Comparison of meiosis and mitosis
Meiosis II - separation of sister chromatids
•
-iosisI-atseparoatichrnoomathoids ogouschromosomes
tosis
(dyads
Me
o f mol
Mi
separ ion f
If we consider the behaviour of genes on chromosomes, alleles segregate during Meiosis I
(separation of homologous chromosomes)
Mendel's ratios predict the segregation of chromosomes during meiosis
(Figure 2-12 part 15)
Demonstration of equal segregation is more straightforward within a haploid life cycle such as in
the yeast S. cerevisiae (Figure 2-17)
Two mating types (MAT and MATa) fuse to form meiocyte
MAT = r+ (white colonies) MATa = r (red colonies)
Figure 2-17 part 2
Meiocyte undergoes meiosis to produce four products (2r, 2r+) - these form colonies directly
Segregation can be followed based on phenotype or by DNA sequence differences (Figure 2-20)
Assigned problems:
9, 17, 20, 21, 22, 26, 28, 32, 36
Ch 2 Questions
29, 31, 38, 40, 46, 48, 52
Meiocyte: undergoes meiosis to produce four products; these form colonies directly
Segregation can be followed based on phenotype or by DNA sequence differences
Probe - a piece of DNA that has a complementary sequence to the DNA of study
Must be long enough
Must be labelled in some way to visualize it
(could be labelled with radioactivity to come up on Xray; flourescent probe; enzyme to
change colour)
genomic DNA - cut with an enzyme (restriction enzymes); cuts at specific sequence
Genotipic Electrolysis
Restriction enzyme cuts at a site along DNA strand
gel in an electric field
put DNA in a hole in the gel; DNA runs down to result in a smear of DNA with large DNA fragments on top
and smaller fragments falling down to bottom of gel
This smear is added to a membrane, where a probe can be used
Forward genetics: Discovering genes controlling flower development
Sex-linked single-gene inheritance patterns
Genes that show different phenotypic ratios in the sexes
Autosome - non sex chromosome
2 copies in all individuals
Sex Chromosomes - sexes have different copies
eg) XX = female; XY = male
Homogametic sex (XX) female in humans and drosophilla
Heterogametic sex (XY) male in humans and drosophilla
Pseudoautosomal region - sequence similarity allows chromosome pairing
Differential region - accounts for sex linkage
Drosophilla
Wild type red eyes = w+
Wild type white eyes = w
Reciprocal Crosses:
Yellow x green and green x yellow will always have same phenotypic results because coloration
is
an autosome
In Mendels' monohybrid crosses..
Reciprocal Crosses (red(f)x white(m) and white(f) x red(m)) give different results for genders
Xw+Xw+ by XwY
red (f) x white (m)
Result: all red eyes
Pedigree analysis: applying Mendelian principles to human inheritance patterns
Characteristics of autosomal dominant pedigree: Trait apprears in every generation
Affected parents have unaffected children
No correlation between sex and a particular phenotype
Chapter 2 - Sex linkage and Pedigrees
Definitions
autosome - a chromosome that is not a sex chromosome
sex linkage - the location of a gene on a sex chromosome
hemizygous - a gene present in only one copy in a diploid organism
Chapter 2 questions (second set):
29, 31, 38, 40, 46, 48, 52
•
Sex-linked single-gene inheritance patterns
genes that show different phenotypic ratios in the sexes
Model Organism: Drosophila, as in humans chromosomal based sex determination, XX (homogametic sex)
= female, XY (heterogametic sex) = male
Human sex chromosomes - X and Y chromosomes
Differential region - accounts for sex linkage (Figure 2-25)
Pseudoautosomal region - sequence similarity allows chromosome pairing
Red-eyed and white-eyed Drosophila (Figure 2-26), isolated by Thomas Morgan, 1910
Wild type = w+ (red eyes)
Mutant = w (white eyes)
In Mendel's monohybrid crosses, reciprocal crosses yielded identical results; in the case of red and
white eyed flies, reciprocal crosses yielded different results
Reciprocal crosses (red x white and white x red) give different F1 results (Figure 2-27)
Reciprocal crosses (red x white and white x red) give different F2 results (Figure 2-27)
Pedigree analysis - applying Mendelian principles to human inheritance patterns (Figure 2-28)
Is the trait shown in generation IV autosomal or sex-linked? Dominant or recessive? (Figure 2-29)
•
•
No tendency of phenotype with a particular sex (note small sample size): suggests autosomal
Unaffected individuals give rise to affected individuals: recessive
If affected individuals in generation IV are homozygous recessive, what genotype must their
parents have been? - parents must be heterozygous in order to produce affected offspring (Figure 229)
Phenotype of siblings = dominant phenotype
Genotype of siblings in generation IV?
•Must have A; may be AA or Aa (indicated as A/-); 2/3 Aa, 1/3 AA
What about genotypes of grandparents (generation II)?
•Both are unaffected, therefore both have dominant allele
•At least one must carry the recessive allele
•If condition is rare, unlikely that both carry recessive allele
Siblings in generations II and III
•Unaffected, therefore must be A
•Other allele is unknown (either A or a); designate as A/In generation III, individuals have 50% chance of being AA, 50% chance of being Aa
Genotypes of Great-grandparents (generation I)?
•Unaffected, therefore must have A allele
•Produce heterozygous progeny, therefore at least one must have allele a, unlikely that both have
allele a
Pseudoachondroplasia phenotype - inherited as an autosomal dominant (Figure 2-30)
•
•
•
Characteristics of autosomal dominant pedigree:
Trait appears in every generation
Affected parents have unaffected children
No correlation between sex and a particular phenotype
If trait is dominant, all unaffected individuals must be homozygous recessive
Affected individuals must have dominant allele (figure 2-31)
Examples of human dominant traits:
Huntington disease: dominant autosomal, late onset means that affected individuals have often had
children before disease manifests (Figure 2-32)
Extra digits (Polydactyly) also a dominant autosomal trait (figure 2-33)
Pedigree of family showing Polydactyly - phenotype of both hands (top L,R) and feet (bottom L,R) indicated
(Figure 2-33b)
•
•
•
X-linked recessive inheritance (Figure 2-36-Figure 2-38)
Phenotypeis more commoninmales thanfemales
Affected fathershave no affected children
Traitpassed from grandfather tograndson
Inheritance in an X-linked dominant disorder (Figure 2-39)
Fathers pass trait to all daughters but no sons
Mothers pass trait to 1/2 sons and daughters
Y-linkage:
Phenotype passed from father to all sons but no daughters
Maleness - primarily determined by SRY gene on the differential region of the Y chromosome
Hairy ears: a phenotype proposed to be Y linked (Figure 2-40)
Clicker Question #1: Which of the following events occurs in Meiosis but not Mitosis?
a) Chromosome condensation
b) Separation of chromatids
c) Pairing of homologous chromosomes
d) Alignment of chromosomes along metaphase plate
e) None of the above
Clicker question #1
In one cross, a female Drosophila with normal bristles is crossed to a male with forked bristles. The F1 has
all normal bristles. In a second cross, a female with forked bristles is crossed to a male with normal
bristles. The F1 is 50% wild type and 50% forked bristles.
The best description of the forked allele is:
a) It is dominant and autosomal
b) It is recessive and autosomal
c) It is recessive and sex-linked
d) It is dominant and sex-linked
CHAPTER 3: Independent Assortment of Genes
First problem set: 1, 2, 6, 8, 9, 17, 19, 46
Second problem set: 13, 28, 37, 39, 41
dihybrid: an individual heterozygous at two genes
product rule: the probability that two independent events occur together is equal to the product of their
individual probabilities
sum rule: the probability that either of two mutually exclusive events occurs is equal to the sum of their
individual probabilities
The green revolution in agriculture
-Rice, wheat and corn that are shorter, disease resistant
-Norman Borlaug - greatest man of the 20th century, awarded Nobel Peace Prize, thought to have saved
lives of a billion people
Mendel's law of independent assortment = Mendel's second law
Mendel studied seven phenotypic pairs (Figure 2-9)
what happens if pairs of traits are considered together?
Mendel crossed round, green seeded individuals (R/r.y/y) by wrinkled yellow individuals (r/r.Y/Y) - "."
indicates relative position of genes is unknown (Figure 3-4)
F1 individuals show dominant traits for both characters - they are heterozygous at both genes
(R/r.Y/y) = dihybrid
Deriving 9:3:3:1 using a Punnett's square
If genes segregate independently of one another, what gametes do we expect and in what frequencies?
(Figure 3-4)
¼R.Y
¼ R.y
¼ r.Y
¼ r.y
Punnett's square predicts that random fusion of four gamete types will give rise to a 9:3:3:1 ratio
Mendel's second law:
During gamete formation, the segregation of alleles of one gene is independent of the segregation of the
alleles of another gene
The product rule predicts a 9:3:3:1 ratio from two independent monohybrid (3:1) ratios
Using a branch diagram to derive 9:3:3:1
Test-crossing a dihybrid
Y/y.R/r x y/y.r/r
If a dihybrid is testcrossed, expect the frequency of phenotypes in the test cross progeny to reflect the
gamete frequency of the dihybrid
Calculating genotypic ratios:
Genotype ratior for F2 of a dihybrid will be (1:2:1 x 1:2:1) = 1:2:1:2:4:2:1:2:1
Calculating individual probabilities:
If only one fusion event results in a particular genotype
Use product rule: the probability of independent events is the product of their individual
probabilities
e.g.
p(yyrr) = 1/4 x 1/4 = 1/16
If more than one fusion event results in a particular genotype
-Apply both product rue and sum rule
-Sum rule: The probability of either of two mutually exclusive events occurring is the sum of their
individual probabilities
e.g. P (YyRr) = [p(YR) x p(yr)] + [p(yr) x p(YR)] +[p(Yr) x p(yR)] + [p(yR) x p(Yr)]
= 1/16 + 1/16 + 1/16 + 1/16
= 1/4
How to calculate the frequency of progeny from a cross where multiple genes are considered?
E.g. in the cross: AaBbCcDdEe x AabbCCddEe what is the probability of an individual with genotype
A_bbC_D_ee)
= 3/4 x 1/2 x 1 x 1/2 x 1/4
= 3/64
The following pedigree concerns a rare disease:
Is the disease causing allele:
a)autosomal, dominant
b)Autosomal recessive
c)Sex-linked recessive
d)X-linked dominant
e)None of the above
What is the probability that A and B will have an affected child?:
a)1/4
b)1/12
c)1/16
d)1/8
e)None of the above
Two dihybrid individuals (A/a B/b) are crossed. If A and B are independently segregating, what is the
probability of progeny that is homozygous recessive at both A and B?
A) 9/16
B) 1/4
C) 3/16
D) 1/16
E) none of the above
Chapter 3 - Independent Assortment of Genes
Mendel's second law predicts independent segregation of chromosomes during anaphase I of meiosis
Interphase: chromosomes are unpaired
Prophase I: chromosomes and centromeres have replicated, but centromeres have not split
Metaphase I: chromosomes align along plate
Anaphase I: centromeres attach to spindle and are pulled to poles of cell
Telophase I: two cells form
AnaphaseII: New spindles form and centromeres finally divide
End of meiosis
Four cells produced frm each meiosis
Recombinants are meiotic output different from meiotic input
Parentals - meiotic output identical to meiotic input
Two ways to form a dihybrid
1
AABB x aabb
AB
ab
P
Meiotic input
F1
or
2
AAbb x aaBB
Ab
aB
AaBb
Meiotic output
¼ AB
¼ab
¼Ab
¼aB
1
parental
"
recombinant
"
2
¼ AB recombinant
¼ ab "
¼ Ab parental
¼ aB "
See textbook
Fig3-12 Recombinant gamete frequency inferred from phenotypic frequency in test cross progeny
In a haploid life cycle (eg. Neurospora) products of single meiocyte demonstrate principle of independent
segregation. Fig 3-9. Eight asci are produced because after meiosis, each gamete undergoes mitosis.
Segregation of homologous chromosomes and sister chromatids are reflected in the arrangement of
spores
If two strains are crossed, two octad types result in equal frequency
Organelle Genomes
The genomes within chloroplasts and mitochonrion
Mitochondrial DNA and chloroplast DNA are relatively small
Contain important genes; ribosomal RNA tRNA and genes controlling cellular energy
Mitochondrial mutation
Poky = slow growth phenotype in Neurospora - if poky females crossed with wide type males, all haploid
progeny show poky phenotype, whereas nuclear gene (ad+) shows a 2:2 segregation
In reciprocal cross (wild type femals crossed to poky males), all haploid progeny show wild type phenotype
of offspring depends on maternal genotype (maternal inheritance)
In reciprocal cross (wild type females crossed to poky males), all haploid progeny show wild type
phenotype, nuclear gene (ad+) still shows 2:2 segregation - phenotype of offspring depends on maternal
genotype (maternal inheritance)
Variegated leaves caused by a mutation in cpDNA (Figure 3-21)
Variegated leaves - mixture of pigments
Due to segragation of organelles within mitotically dividing organism
Some branches all white
Some branches all green
Main shoot and some branches both green and white = variegated
(eg) this can be mutant and wild chloroplasts
Cytoplasmic inheritance - phenotype of offspring depends on genotype of maternal cytoplasm (white,
green, or variegated)
Crosses using flowers from a variegated plant (Figure 3-22) - phenotype of offspring depends on
cytoplasmic genotype of mother (cytoplasmic inheritance)
Sites of mtDNA mutations in certain human diseases
e.g. MERRF (mycoclonic epilepsy and ragged red fibre) single base change in mtDNA (Figure 3-24)
Typical maternal inheritance in a human pedigree (Figure 3-25)
Chi-Square (2) Test
Use a statistical test to determine if the observed results are significantly different from the expected
e.g. in a test cross AaBb x aabb
1) develop null hypothesis
The genes, A and B, are independently assorting (i.e. they are not linked) and the test cross progeny do not
deviate from the expected 1:1:1:1 ratio
Calculate 2 value
X2 = sum of (observed-expected)2/expected =  (o - e)2/e
degrees of freedom (df) = number of classes (n) -1
=3
compare the calculated chi-square value to the critical value from the chi-square table (Table 3-1)
-if
-if
2
is < critical value, P is > than 0.05, accept null hypothesis
2
is > critical value, P is < than 0.05, reject null hypothesis
100 > 7.815, reject hypothesis
Clicker question 1:
The following cross is done:
Aa;bb;cc;DD;Ee;Ff;Gg x aa;Bb;CC;Dd Ee;ff;Gg
In the progeny, what is the probability of an individual having the phenotype
A_ bb C_ D_ ee ff G_?
A) 0
B) 3/64
C) 3/128
D) 1/128
E) none of the above
CHAPTER 4 - Mapping Eukaryote Chromosomes by Recombination
Question set 1: 1, 2, 3, 6, 7, 8, 10, 12, 13, 16
Question set 2: 21, 22, 24, 25, 36, 37, 42, 50, 51, 53, 56
recombinant product: a product of meiosis that differs from the haploid genotypes that gave rise to the
diploid meiocyte
parental product: a product of meiosis that is the same as the haploid genotypes that gave rise to the
diploid meiocyte
dihybrid: an individual heterozygous at two genes
coupling configuration - arrangement of linked alleles in a dihybrid such that the two dominant alleles
are on one chromosome, the two recessive alleles are on the other chromosome (AB/ab)
cis dihybrid - a dihybrid in which the alleles are coupled (AB/ab)
repulsion configuration - arrangement of linked alleles in a dihybrid such that the a dominant and a
recessive allele are together on each chromosome (Ab/aB)
trans dihybrid - a dihybrid in which the alleles are in repulsion (Ab/aB)
map unit (centimorgan, cM): the distance between two genes where one product of meiosis out of 100 is
recombinant
gene locus: the position of a gene within the genome
linked - two genes that are on the same chromosome
linkage group - a set of genes that are on the same chromosome
William Bateson and Reginald Punnett
Sweet Pea
P - gene for flower colour
L - gene for seed shape
Expected independent segregation of P and L meaning:
1:1:1:1 ratio of dihybrid test cross progeny
9:3:3:1 ratio of dihybrid F2 progeny
Instead, following selfing of a dihybrid, ratio was quite different from 9:3:3:1 and Bateson and Punnett
observed:
P and L are coupled
Dominant alleles stay together and recessive alleles stay together
Larger classes are those where P is together with L, or p together with l
= configuration of alleles in parents
Smaller classes are those where P is together with l, or p with L
= different configuration from parents
Thomas Morgan - Drosophila
Eye colour - pr = purple,
Wing shape - vg = vestigial,
pr+ = red,
where pr + > pr
vg + = normal, where vg + >vg
Like Bateson and Punnett, Morgan observed configuration of alleles in parental lines is more common
in F2 progeny
Hypothesis - pr and vg are on the same chromosome
Experiment - test cross
First cross:
Parental cross: pr+/pr+.vg+/vg+ x pr/pr.vg/vg
F1 progeny: pr+/pr.vg+/vg
If genes assort independently, what phenotypic ratio do you expect in the test-cross progeny?
Expected result is 1:1:1:1
Results: Configuration of alleles in parental lines is more common in test cross progeny
Dominants are together, recessives together = coupling = cis (adjacent) conformation
Annotation: pr+ vg+ /pr vg
Second cross:
Parental cross: pr+/pr+.vg/vg x pr/pr.vg+/vg+
F1 progeny: pr+/pr.vg+/vg
Results: Configuration of alleles in parental lines is more common in test cross progeny
Alleles are in repulsion = trans (opposite) configuration = trans dihybrid
Annotation: pr+ vg /pr vg+
Conclusion: alleles that are together in the parents are more likely to stay together through subsequent
generations
Independent of dominance/recessive
Conclusion: genes are linked (on the same chromosome)
Linked alleles tend to be inherited together (figure 4-2)
Morgan suggested that crossing over produces allelic combinations that are different from the meiotic
input (recombinants) (Figure 4-3)
Morgan suggested that chromosomes break and re-join during pairing, and that chiasmata are
visual evidence of that event
Chiasmata in grasshopper testes (Figure 4-4)
4.1 CROSSING OVER OCCURS DURING PROPHASE
Morgan proposed that crossing over is between chromatids, not chromosomes (Figure 4-5)
A second-division segregation pattern in a fungal octad demonstrates that crossing over must
be between chromatids (Figure 4-21)
Multiple crossovers can include more than two chromatids to generate multiple types of
linear tetrads (figure 4-6)
SUMMARY: Crossovers generate recombinant meiotic products (Figure 4-7)
4.2 MAPPING BY RECOMBINANT FREQUENCY
Morgan's results: If test crosses were done involving different pairs of genes, saw different ratios
of parental: recombinant
Morgan's hypothesis: frequency of recombinant products may be related to distance between
genes
Alfred Sturtevant: If Morgan's hypothesis is true, should be able to use recombinant frequency to
establish spatial relationship between genes on chromosomes - i.e. a chromosome map
Alfred Sturtevant - data from Morgan's lab, generated formula
map distance = percent recombinants
Predicted that if calculated distance reflect linear position on chromosome, map distances should
be additive (Figure 4-9)
Sturtevant's map of the Drosophila X-chromosome
Longer regions have more crossovers and thus higher recombinant frequencies (Figure 4-10)
For linked genes, recombinant frequencies are less than 50 percent (Figure 4-8)
(because it is more difficult for genes to separately cross over when they are positioned one on top of the
other on a chromatid)
If two genes are more than 50 cM apart on the same chromosome, cross-overs will occur so frequently that
the genes will assort independently, therefore recombination frequency never exceeds 50%
For these genes, we can only detect that they are on the same chromosome by showing that each is linked
to a common gene
Knowing recombinant frequency, we can calculate map distance:
Map distance (mu or cM) = % recombinant products
Knowing the map distance between genes, we can predict the frequency of recombinant products in
a cross
Calculated map distance is correlated to physical distance on DNA molecule - knowing the position
of the gene in the chromosome allows us to find the sequence of gene and hence better understand
its structure and function
Clicker question #1
In Arabidopsis, seeds with the dominant TT8 allele are brown, whereas those with recessive tt8 allele are
yellow; seedlings with dominant HY4 alelle have short hypocotyls, while those with recessive hy4 have
long hypocotyls. A test cross is performed on dihybrid individual and the following progeny are obtained:
TT8 HY4
24
TT8 hy4
178
tt8 HY4172
tt8 hy4
26
The best description of the dihybrid is:
a) TT8/tt8.HY4/hy4
b) TT8/tt8;HY4/hy4
c) TT8 HY4/tt8 hy4
d) TT8 hy4/ tt8 HY4
e) None of the above
Clicker question #2
A maize plant is dihybrid for two genes BR and DW. This dihybrid individual (BR/br DW/dw) is testcrossed, and progeny of the following phenotypes are seen:
BR DW 240
br dw
234
br DW
64
BR dw
72
The best description of the genes in the dihybrid is:
a) they are coupled
b) they are sex-linked
c) they are independently assorting
d) none of the above
Clicker question #3
In Arabidopsis, seeds with the dominant TTG allele are brown, whereas those with recessive ttg allele are
yellow; seedlings with dominant HY4 alelle have short hypocotyls, while those with recessive hy4 have
long hypocotyls. A test cross is performed on dihybrid individual and the following progeny are obtained:
TTG HY4
24
TTG hy4
178
ttg HY4172
ttg hy4
26
The best description of the dihybrid is:
a) TTG/ttg.HY4/hy4
b) TTG/ttg;HY4/hy4
c) TTG HY4/ttg hy4
d) TTG hy4/ ttg HY4
e) None of the above
Clicker question #4
In Arabidopsis, seeds with the dominant TTG allele are brown, whereas those with recessive ttg allele are
yellow; seedlings with dominant HY4 alelle have short hypocotyls, while those with recessive hy4 have
long hypocotyls. A test cross is performed on dihybrid individual and the following progeny are obtained:
TTG HY4
178
TTG hy4
24
ttg HY426
ttg hy4 172
The truebreeding parents of the dihybrid must have had genotypes
a) TTG TTG hy4 hy4 x ttg ttg HY4 HY4
b) TTG TTG HY4 HY4 x ttg ttg hy4 hy4
c) TTG ttg HY4 hy4 x TTG ttg HY4 hy4
d) TTG HY4 x ttg hy4
e) None of the above
Chapter 4 - Mapping Eukaryote Chromosomes by Recombination
Questions: 1, 2, 3, 6, 6, 7, 10, 11, 12, 13, 16
Sweet Peas
P - gene for flower colour
L - gene for seed shape
Expected segregation of P and L
1:1:1:1 ratio of dihybrid test cross
9:3:3:1 dihybrid progeny
Cross:
Purple Long
PPLL
F1:
PpLl
F2:
9 P_L_
3
3
1
X
x
red round
ppll
(purple, long)
Expect
(215)
Observe
(284)
increased
decreased
decreased
increased
William Bateson and Reginald Punnett observed:
Larger classes are those where P is together with L or when p with l
= configuration of alleles in parents
Smaller classes are those where P is together with l, or when p with L
= different configuration from parents
Coupling: dominants together and the recessive together
Thomas Morgan: Drosophila
Nobel Prize 1933
Eye colour - pr = purple
pr+ = red, pr+ > pr
Wing shape - vg = vestigial
vg+ = normal, vg+ > vg
Like Bateson and Punnett, Morgan observed configuration of alleles in parental lines is more common in F2
progeny
Hypothesis: pr and vg are on the same chromosome
Morgan: If test crosses were done involving different pairs of genes, saw different ratios of parental :
recombinant
The frequency of recombinant products may be related to distance between genes
Independently assorting results in
1:1 : 1:1
if they are liked then the ratio changes
P increase : P decrease
Alfred Sturtevant said that if this is true, should be able to use recombinant frequency to establish spatial
relationship between genes on chromosomes - ie. a chromosome map
Predicted that if calculated distance reflect linear position on chromosome, map distances should be
additive
% recombinant products = map unit (called m.u. or centiMorgans or cM)
Transdihibrid
AaBb x aabb
95 Ab
95 aB
6 ab
4 AB
Parental phenotype
"
Recombinant
Recombinant
Map Distance = observed (1) + observed (2) x 100%
total products
Longer regions have more crossovers and thus higher recombinant frequencies
see txt Fig 4-10
independently assorting
50% recombination frequency
for linked genes, recombinant frequencies are always less than 50%
if two genes are more than 50cM apart on the same chromosome, recombinations will occur so frequently
that the genes will assort independently, therefore recombination frequency never exceeds 50 %
Knowing recombinant frequency, we can calculate map distance:
map distance (mu or cM) = %recombinant products
Knowing the map distance between genes, we can predict the frequency of recombinant products in a cross
Transdihibrid for A_C_ is test crossed. What is the frequency of "ac" phenotypes in the progeny?
Ac / aC A
c
a
C
we would expect that if they are 20 mu apart;
if the recombinant products (Ac/aC) are 20% then we want half of those for
"ac" so the percent is 10% for "ac"
Transdihibrid is selfed
What is the frequency of "ac" phenotypes on F2? The allele "a" is recessive.
Parentals: need specifically from mother and father
Female "ac" x Male "ac"
Aacc = (Ac)(ac) + (ac)(Ac)
= 0.1 x 0.1
= 0.01
Therefore the frequency is 1%
Calculated map distance is correlated to physical distance on DNA molecule - knowing the position of the
gene in the chromosome allows us to find.
CHAPTER 4 - Mapping Eukaryote Chromosomes by Recombination
Question set i: 1, 2, 3, 6, 7, 8, 10, 12, 13, 16
Question set ii: 21, 22, 24, 25, 36, 37, 42, 50, 51, 53, 56
trihybrid: an individual heterozygous at three genes
interference: a measurement of the degree to which a cross-over in one interval inhibits a cross-over in a
second interval
THREE POINT TEST-CROSS
Figuring out the distance between 3 genes
Mapping the distance between three genes on the same chromosome
LFY - flowers
lfy - leaves
CER3+ - epidermal waxYI - green inflorescence
cer3 - no wax
yi - yellow inflorescence
Following test crossing to a trihybrid individual, the following phenotypes were scored in the progeny:
450
lfy CER3 YI
462
LFY cer3 yi
1
lfy cer3 yi
1
LFY CER3
26
YI
32
lfy CER3 yi
17
LFY cer3 YI
11
LFY CER3
yi
lfy cer3 YI
From the data, we know that the true-breeding parents that were crossed to generate the F1:
Parental cross: lfy CER3 YI x LFY cer3 yi
F1: lfy CER3 YI / LFY cer3 yi
METHOD 1 - Mapping 3 genes by calculating 3 pairwise distances:
1. Consider LFY- CER3 distance (ignore YI)
distance between LFY and CER3 = % recombinants between LFY and CER3
Distance LFY and CER3 = ( lfycer3yi + LFYCER3YI + LFY CER3yi + lfycer3YI)/1000
recombinant frequency = (1 + 1 + 11 + 17)/1000
= 0.03
=3%
map distance = 3.0 cM
2. Consider CER3 - YI distance (ignore LFY)
Distance between CER3 and YI = % recombinants between CER3 and YI
recombinant frequency = (26 + 32 + 11 + 17)/1000
= 0.086
= 8.6%
map distance = 8.6 cM
3. Distance between LFY and YI (ignore CER3)
distance between LFY and YI = % recombinants between LFY and YI
recombinant frequency = (1 + 1+ 26 + 32)/1000
= 0.06
=6%
map distance = 6 cM
[----------------------- (8.6) ------------------------]
CER3
LFY
YI
[-------- 3 --------][--------------- 6 ---------------]
Notice that 3 + 6 is not 9 in this case, therefore we made a calculation error.
We missed a recombinance when we calculated the distance between CER3 and YI.
Therefore we counted something as parental instead of recombinance.
Distance CER3 and YI = 26 + 32 + 17 + 11 + 2(1+1) x 100%
1000
= 9 cM
We can avoid this problem by only calculating the two smaller distances.
Next time we can just figure out that LFY is in the middle, and just calculate the distances from CER3 to LFY
and then the LFY to YI.
We recognize that LFY is in the middle because its alleles switch places relative to the outer two genes in
the DCO class. Notice how the CER3 and YI stay same in the biggest two classes which are the
parentals.
Parental
Parental
CER3 lfy
cer3 LFY
YI
yi
Change:
Change:
CER3 LFY
cer3 lf y
YI
yi
Most likely arrangement based on distances:
cer3 and yi are farthest apart
Most accurate measurement of cer3 - yi distance is the sum of the other two pairs
3 cM + 6 cM = 9 cM
So, why did we calculate a map distance of 8.6 cM?
When we calculated distance between CER3 and YI, we considered lfy cer3 yi and LFY CER3
YI
parentals
In fact, these are double crossover (DCO) classes, resulting from one cross-over between YI and
LFY
and one cross-over between LFY and CER3
By comparing the parental (Largest) classes to the DCO (smallest) classes, we can recognize
the gene in the middle, because its alleles switch places relative to the outer two genes in the
DCO class
Rules for a three point test cross:
1) there should be 8 phenotypic classes
2) progeny classes are grouped in pairs = reciprocal cross over events
3) largest classes are the parentals
4) smallest classes are the double cross-over classes (DCO)
Recombinant frequency between outer genes
= sum of frequency between two pairs
= (sum of SCO classes) +2(sum of DCO classes)
total
= (26+32+11+18) + 2(1+1)
1000
= 0.09
Map distance = 9 cM
Calculating interference:
A measurement of the independence of two cross-over events (does one cross-over interfere
with
the possibility of another cross-over)
If crossovers are independent of each other, frequency of two cross-overs should be the
product of their individual probabilities
Interference
= 1 - coefficient of coincidence = 1 - observed DCO
expected DCO
= 1 - observed Double Cross Overs
p(SCOI) x p(SCOII) x total progeny
Interference = 1 -
observed DCO
p(SCOI) x p(SCOII) x total progeny
Example lfy cer3 yi:
Observed DCO = 2
Expected DCO = (0.06)(0.03)1000
=1.8
Interference
=1 - 2/1.8
= - 0.11
Negative interference - more double-crossovers than expected (cross-over in one region promotes crossovers in another region)
Positive interference (usually)- fewer double-crossovers than expected (cross-over in one region inhibits
cross-overs in another region)
Clicker question #1
In Arabidopsis, seeds with the dominant TTG allele are brown, whereas those with recessive ttg allele are
yellow; seedlings with dominant HY4 alelle have short hypocotyls, while those with recessive hy4 have
long hypocotyls. A test cross is performed on dihybrid individual and the following progeny are obtained:
TTG HY4
24
TTG hy4
178
ttg HY4172
ttg hy4
26
The best description of the position of TTG and HY4 is:
a) they are on different chromosomes
b) They are more than 50 cM apart on the same chromosome
c) They are 12.5 cM apart on the same chromosome
d) None of the above
BECAUSE: If a gene segregates independently; a test cross to the dihybrid will give a 1:1:1:1 ratio. Then,
the alleles will be either more than 50cM apart on the same chromosome or they are on different
chromosomes.
Map distance = %recombinance
TTG HY4
24
TTG hy4
178
ttg HY4172
ttg hy4 26
map distance between TTG and HY4 alleles = ( 24+26 ) x 100%
400
= 12.5 map units apart
Clicker question #2
In tomato, genes R and WF are 15 cM apart on Chromosome 2. Gene R controls fruit colour (dominant allele
R results in red fruit, recessive allele r in yellow fruit) while gene WF controls flower colour (dominant
allele Wf results in yellow flowers, recessive allele wf in white flowers)
An plant that is cis-dihybrid for R and WF is test-crossed. In the test cross progeny, what will the frequency
of yellow fruit and white flowered plants be?
a)85%
b)42.5%
c)15%
d)7.5%
e)None of the above
ANSWER: We are looking for a parental class and since it is a cis-dihybrid with double recessive
Map distance = % recombinance
(b) 42.5%
Three Point Test Cross Examples
CLICKER 3
An individual trihybrid at three genes P, Y, and T is test-crossed for 500 progeny.
PYt
144
pyT
141
PYT
7
pyt
8
Pyt
24
pYT
26
pYt
70
pyT
80
We can see that P and Y pair frequently, so T is in the middle on the chromosome.
YT distance = [ ( 24+ 26 + 7 + 8 ) / 500] x 100% = 13 % = 13 cM apart
TOTAL:PY distance = [ ( 24 + 26 + 70 + 80 + 2(7+8) ) .. 46
Clicker question 4
Within 1000 progeny
bTS
btS
BTS
BtS
bts
bTs
BTs
Bts
14
48
355
83
350
87
50
13
The 'B' allele is in the middle: TBS or SBT
Find the distance between B and S where map distance is the per cent
recombinance
B-S distance = (14 + 48 + 50 + 13) x 100%
= 12.5 cM
1000
Clicker Question 5
Test Cross Progeny
ABC 450
abc
440
Abc
4
aBC
6
ABc
25
abC
30
aBc
25
AbC
20
The A allele is in the middle such that BAC or bac are parentals
A-B distance = (4 + 6+ 25 + 20) x 100%
1000
= 5.5 cM
A-C distance = (4 + 6 + 25 + 30) x 100%
1000
= 6.5 cM
B-C distance = 5.5 cM + 6.5 cM
[---------------12 cM-------------]
= 12 cM
Map
B
[
5.5
A
][
interference = 1 - observed DCO
expect DCO
6.5
C
]
= 1 - ( 4 + 6)
(0.055)(0.065) x 100
= -1.8 interference
A trihybrid individual with Bad / bAD is test crossed. What is the probability of bad in the progeny?
B
A
D
20
30
In order for this to happen, a cross over must occur between A and B. What is the probability of this
happening?
B
b
a
A
d
D
20 cM = % recombinance of single combination
30 cM = % recombinance or cross over of A and D
No recombinance is 70% or 0.7 between A and D
p(bad) = ½ p(SingleCrossOverI) (NCO II)
= ½ (0.2) (0.7)
= 0.07
=7%
* we make ½ apart of this probability because when B/b cross
it causes two results (bad as well as BAD) we only want one
What happens when Bad/bAD is selfed? What is the probability of bad in progeny?
p(bad) = [ ½ p(SCOI)(NCOII) ]2
= [ ½ (0.2) (0.7) ]2
= 0.49 %
Chapter 4 questions (set 1): 1, 2, 3, 6, 7, 8, 10, 12, 13, 16
Questions (set 2): 21, 22, 24, 25, 36, 37, 42, 50, 51, 53, 56
4.4 Centromere mapping with Linear Tetrads
Chromosomes within spores of octad reflect position of chromatids during meiosis
During two divisions:
Division 1 - separation of homologues
Division 2 - separation of sister chromatids
1st division (MI) segregation: meiosis in a heterozygote meiocyte; different alleles separate after MI - a
4:4 ratio (order +number)
Indicates that 2 chromatids in each homologue were identical
indicates that no crossover had occurred between gene and centromere
2nd division (MII) segregation: meiosis in a heterozygote meiocyte; different alleles separate after MII
(2:2:2:2 ratio)
Indicates that 2 chromatids linked by the centromere were not identical
a cross-over had occurred between the gene and the centromere
4: 4
MI segregation pattern
No recombinance
2:2:2:2
MII segregation pattern
½ MII spores are recombinants
Distance = % recombinant products
-MII segregation pattern indicates that a cross-over occurred within a meiocyte
-of the spores produced by that meiocyte, ½ are recombinant
Map Distance = % recombinant products
= ½ MII segregation pattern frequency
= ½ (MII octads / total )x 100
In linear octads, frequency of MII segregation (2:2:2:2) is a phenotypic indication of recombination
between a gene and the centromere.
We can use this phenotypic frequency to calculate the distance between the gene and the centromere The
further the gene is from the centromere, the more frequently that MII segregation will occur.
Example:
Half the spores are recombinant and half parental in the MII asci
Experiment: a diploid Sordaria, of genotype Aa undergoes meiosis
Results: 1000 linear octads are scored as MI (4:4) or MII (2:2:2:2) patterns
Distance A to centromere = ½(MII tetrads) x 100
1000
= ½ (13+9+12+11) x 100
1000
= 2.25 m.u.
Map distance = % recombinants
= [½ MII asci]/total x 100%
*only situation where the one half ½ fraction is used
Four different spindle attachments produce four second-division segregation patterns (Figure 4-22)
4.3 Mapping with Molecular markers
Molecular markers:
DNA sequence variation between individuals
o May have no phenotypic consequence
o Much more common than traditional phenotypic based markers
e.g. in human genome, 1 SNP/1000 bases
o Molecular heterozygotes can be identified
o Can identify one recombinant, whereas if the phenotypic marker is a recessive
allele, two recombinants must combine to be seen
o Test crosses are not as critical
Examples of Molecular markers:
Single nucleotide polymorphisms (SNPs)
o Silent in genes or intergenic regions
o Causing a mutant phenotype
Simple sequence length polymorphisms (SSLPs)
o Mini- and micro-satellite markers
Phenotypic and molecular markers mapped on human chromosome 1 (Figure 4-20)
those highlighted in green are associated with a phenotype
purple = SSLP
blue = other polymorphism
green = gene, other = intergenic region
Detecting SNPs
By sequencing - not always possible since this requires prior knowledge of sequence and is
expensive
By RFLP (restriction fragment length polymorphism) analysis
By CAPS (cleaved amplified polymorphic sequence) analysis
Restriction fragments: fragments generated by restriction ezymes/ endonucleases
Nucleases: enzymes that cut nucleic acids / DNA and RNA
Ectonuclease
Endonuclease
RFLPs
Restriction enzymes - cleave DNA at a particular sequence (e.g. EcoRI cuts at
5'-GAATTC-3' 3'CTTAAG-5'
If mutation changes sequence (AT to GC):
5'-GAGTTC-3'
3'-CTTCAG-5'
Restriction enzyme is no longer able to cut
IF a SNP changes a restriction enzyme recognition site, one sequence variant can be cut, the other
cannot = RFLP
RFLP is detected by a Southern blot
Performing a Southern blot:
1. separate genomic DNA cut with restriction enzyme via electrophoresis (Figure 20-12)
2. DNA from gel is transferred to a membrane (Southern blotting, Figure 20-12)
3. the membrane (filter) is exposed to a labeled probe complementary to the region of interest (i.e.
containing the RFLP) (Southern hybridization)
4. Bound, radioactively labeled probe will expose an x-ray film, and identify DNA fragments to which
probe has bound
SNP - single nucleotide polymorphism
Detected by RFLP and CAPS (cleaved amplified polymorphic site)
Amplify the DNA using PCR to isolate DNA before gel electrophoresis
"amplified" - can isolate the DNA using polymerase chain reaction using primers
Primers - provide beginnning and end points for DNA replication
Results will be the same:
- Allele 1 will be amplified
- Allele 2 will be amplified with two band in the gel
Less dangerous: no radioactivity
An RFLP linked to a mouse disease gene (Figure 4-15a)
2 polymorphisms
DNA morph1 = no RE cut site, one fragment recognized by probe
DNA morph2 = RE site, two fragments recognized by probe
Following a test cross, follow segregation of RFLP in individuals with disease (Dd) and those
without (dd) (Figure 4-15b)
morph2 (2 bands) is always associated with disease allele (D) except in the 8th individual
In test cross progeny #8, a cross-over occurred between D and marker (Figure 4-15c)
of 16 chromatids analyzed, one was recombinant, distance = 6.25 m.u.
Using haplotypes (a set of linked markers) to deduce gene position (Figure 4-16)
-closely linked markers will show linkage disequilibrium
linkage disequilibrium: .
-flanking markers = markers 4 and 5 do not dissociate from the mutation
determined because mutant allele/gene of interest will not recombine with 4 and 5 although it does
recombine with all other genes; therefore we deduce that our gene of interest lies between 4 and 5
• CAPS
-primers flanking a polymorphic restriction enzyme site are used to amplify region by polymerase chain
reaction (PCR)
-amplified region is cut by restriction enzymes
-compared to RFLP: avoids use of Southern blot, probes, requires less DNA
SSLPs: Single sequence like polymorphism
1. Minisatellites
Genomes contain large amount of repetitive DNA interspersed through genome
Short sequences of repetitive DNA (15-100 nucleotides) = minisatellites
subject to deletions and expansions
Variable numbers of repeats = different alleles
If DNA polymerase slips, deletion or expansion can occur where different alleles repeat
Detection: cut with a restriction enzyme OUTSIDE the minisatellite (repeated area),
use probe that recognizes repeat - see different sized bands on Southern blot
Minisatellite regions can be used to generate a DNA fingerprint (Figure 4-18)
2. Microsatellites
Very short sequences of repetitive DNA (2nt) = microsatellites
subject to deletions and expansions
Variable numbers of repeats = different alleles
Detection: amplify by PCR using primers that bind to region flanking microsatellite - different size products
reveal different alleles (Figure 4-19)
Gel Electrophoresis
Performing
Readings:
Omit: 4.5, 4.6, 4.7, 4.8
section 4.5 = chi-square to test for linkage, accounts for possible difference in gamete viability
if you are asked to do this, simply use 1:1:1:1 ratio as the expected (null hypothesis = genes are
independent of one another)
Clicker question:
•Three genes (TSB) are arranged on the chromosome with S in the middle, T 5 cM to one side and B 10 cM
to the other side. An individual of genotype tSB/Tsb is test-crossed. What will the frequency of the tsb
phenotype be in the progeny?
•A) .05
•B) .1
•C) .025
•D) .0225
•E) none of the above
Crossover between T and S
t
T
S
s
B
b
5cM
10cM
Chance of a crossover to form tsb
= ½ P(S (O TS) )
x P(Nco(SB))
= ½ (0.05) (0.9)
= 0.225
2-3pm Wednesday office: D8 Tony Russel
Chapter 5: The Genetics of Bacteria and their Viruses
Bacteria and Viruses do have the ability to exchange DNA and undergo recombination. In both types, genes
transfers occur:
Bacteriophage inject DNA into bacteria
While phage is replicating it acquires some host DNA in recombination
Why is it important?
The fruits of DNA technology mad possible by bacterial genetics
-modern gene cloning techniques
at 3 billion base pairs the first complete human genome was sequenced by cloning smaller sections of the
genome
Dividing bacterial cells
-binary fission
first make an identical copy of genome by DNA synthesis phase
from single cell, production of two daughter cells
-rapid exponential growth
Mutation
spontaneous mutation results in two different daughter cells
recombination is a separate process that is NOT dependent on cell division
Separation and visualization:
bacterial cells with different genotypes and phenotypes
It is easier to infer the genotype in bacteria because they are haploid
bacterial cells are grown on plates
each colony is a clone of cells with the same genotype
Process:
Suspension of bacterial cells is spread on a petri plate with agar gel and incubated from 1 to 2 days.
Visible colonies are each a clone of the corresponding single cell from the original suspension. By
looking at the phenotypes present on the plate, we infer the genotypes in the original liquid culture.
Properties of Bacteria
Most are prototrophic: self sufficient, ability to synthesize all essential molecules it requires for growth,
can be grown on minimal media (carbon source and salts)
Auxotrophic mutants require additional nutrient supplements (mutations affect the synthesis of a
molecule such as amino acids or vitamins)
The loss of an ability is denoted with a minus ( — ) sign
Resistance and sensitivity can be denoted by (r) and (s)
Eg) Symbol
Character/ Phenotype
arg requires arginine added as a supplement to minimal medium
lac cannot utilize lactose as a carbon source
strr
resistant to the antibiotic streptomycin (mutant)
sensitive to the antibiotic steptomycin (wild type)
str s
Bacterial Selection
Selecting for auxotrophic:
We can take an auxotrophic mutant into a suspension containing adenine and separate those which can
survive on their own and grow
Properties:
-single - celled
Prokaryotic (no nucleus or membrane-bound organelles)
Have only 1 circular chromosomes
Haploid organisms
Do not undergo meiosis like eukaryotes (where recombination occurs in eukaryotes)
Still undergo recombination
Lederberg and Tatum (1946)
Discovered gene transfer in E. coli
Mixing bacterial genotypes produces rar recombinants
A
B
Strain A is an auxotroph for 2 alleles Strain B is wild for what A is mutant
Met- bio- thr+ leu+ thi+
met+ bio+ thr- leu- thiA+
BMixture
Wild Type
culture
No spontaneous reversion
Instead, need to mix the two in order for colonies to be produced
Two strains, both auxotrophic
Strain A = bio- metStrain B = thr- leu- thiPrototrophs can be detected at a low frequency ~1 in 10 million cells
Is this recombination or do they cross feed from metabolic pathways?
Answer: recombination
Proof: U-tube experiment
U-tube experiment
Bernard Davis separated the two cultures with a filter to keep Strain A on the left and Strain B on the right
with no ability for cell interaction or gene transfer, however molecules can move back and forth
= no recombinants are produced without cell contact
Bacteria conjugate by using pili
Bacteria exchange DNA
Only one we will look at that requires cell contact for gene transfer
Plasmids - extrachromosomal elements
- under certain conditions are non-essential
Uni-directional: middle cell donates while other cells accept:
Conjugation:
Plasmid transfer
during conjugation
<--
Plasmids
Genome
Conjugation
Partial genome tranfer
during conjugation
Question: is the transfer of genetic information bi-directional?

Experiment
1. Strain A (Strs met- leu- bio+ thi+) X Strain B (StrR met+ leu+ bio- thi-)
Allow conjugation
Streptomycin to kill strain A
Result: Prototrophs obtained
2. Stain A (StrR met- leu- bio+ thi+) X Strain B (StrR met+ leu+ bio- thi-)
Conjugation
Streptomycin to kill strain B
Result: No proptotrophs obtained
Conclusion: uni directional
Only the strain B cells were able to produce prototrophs, therefore it is unidirectional from A to B
.
Conjugation
Extra-Chromosomal DNA Plasmid
- one member of a conjugating pair carries the fertility factor (F) on the F plasmid
- F plasmid = small circular, extrachromosomal DNA
- F+ strains: donor strain bacteria contains the F-factor
- F- strains: acceptor that does not contain the F-factor
Properties of F-factor (plasmid)
- enables the production of pili (proteinaceous attachment tube between cells)
- able to replicate during cell division and during conjugation
- prevents conjugation between F+ cells
- only get conjugation between F+ to F- cells
F plasmids transfer during conjugation
Donor F+ cell contains an F plasmid which can produce a pili
Donor cell pulls acceptor cell with pili
Cut in one of the two strands of DNA of F plasmid
One DNA strand is pulled into recipient cell
- transfer of single-stranded DNA
- Replication in both donor and recipient
- Rollin circle replication in donor
The accepting cell will replicate the DNA as it accepts the new DNA
Result: Donor and Recipient both have doublestranded F plasmid
ICLICKER
During Bacterial conjugation, which one of the following would not be expected to occur?
A) an F- cell converted to F+
B) An F+ cell would generate pili to attach to F- cells
C) DNA would be replicated in the F+ cell
D) DNA would be replicated in the F- cell
E) An F+ cell would be converted to FCavalli-Sforza
Observations
1. Occaisionally F+ strains become strains with a high frequency of recombination: Hfr strains
-1000x more frequent transfer of genes than regular F+ strains
2. Hfr x F-only rarely is F- converted to HFR or F+
Hfr can convert auxotrophs to prototrophs:
In Hfr strains, F plasmid has been inserted into the bacterial chromosome
IS = Insertion sequence
The IS is similar to the bacterial chromosome in DNA composition
The IS enables the F plasmid to be added to the bacterial chromosome
The process, then, is reversible
Integration of the F plasmid creates an Hfr strain
F+ with an F plasmid in the cell Hfr with an integrated F plasmid
Crossovers integrate parts of the transferred donor fragment
- Cleavage in the Hfr DNA sequence transfers the single-stranded DNA copy
- Exconjugant is formed with two homologous regions where we can get recombinantion; transferred
fragment is converted into the double helix through DNA synthesis; however not all exconjugants will
undergo recombination
- Recombinant cell is formed where double crossover inserts donor DNA
Exogenote: DNA coming from the donor
Endogenote: DNA originally in the bacterial chromosome
Summary:
1. Mechanism of DNA transfer is the same as F+ x Fa. replication in donor and host
b. single-strand of NDA
2. A portion of Hfr genome gets transferred
a. High recombination frequency
3. Recombination
a. Exogenote 
b. Endogenote ..
Two types of DNA transfer can take place during conjugation
Chromosome transfer vs. Plasmid Transfer
Start: F+ cell with a+ bacterial chromosome
(1) Plasmid: conjugation and transfer of the F factor
Recipient cell will have no chance to replace the a+ allele/ genotype
(2) If there is enough Chromosome transfer then with enough transfer, can integrate the a+ allele
but not the F plasmid
Interrupted Mating Experiment
Wolman and Jacob combined:
Hfr strain: O thr+ leu+ azir tonr lac+ gal+ strS
F- strain: thr- leu- aziS tonS lac- gal- strr
Plate: media with streptomycin (kills Hfr) and no theonine or leucine provided (selects for recombinants
that underwent conjugation)
Then, scored for the underlined genotypes
Tracking time of marker entry generates a chromosome map
Frequency of Hfr genetic character among strr exconjugates vs. time
Gives us a sense of distance between the alleles on the chromosomal DNA
Origin is a specifice site where DNA cleavage and DNA replication begin
Alleles that were closest to the origin are more likely to be transferred
Time of entry is dependent on position relative to F plasmid insertion
The first allele appearing in the new cell shows its close distance to origin
The time taken to transfer is related to the distance between the gene and origin
Therefore, we obtain a crude chromosome map of genes
Questions:
Is the relative position of genes, the position of the origin, and the direction of transger the same in
different Hfr strains?
Experiment:
Consider 5 different Hfr strains derived.
Example
Isolated 5 different Hfr strains
We infer the site of integration to be different for each strain
Relative order between alleles in the strains are different
Relative position of the alleles, however, are the same (eg) b is always associated with a or c on either side
Results:
1) The first gene to enter may differ
2) The relative position of each marker (gene) is constant
3) The overall directionality of transfer can differ
Conclusion: The difference between the Hfr strains is the relative position and orientation of the origin
within a circular chromosome.
This experiment proved that E.coli has circular DNA molecule
ORIGIN (O) -- A
B
C
D
E -The only way to explain the transfer is if A and E are joined in a circle
What controls these insertion patterns?
Why are there different insertion sites?
There are multiple insertion sequences (IS) in the E. coli genome.
A single crossover inserts F at a specific locus
Orientation of IS determines the order of gene transfer
Homologous regions where pairing can take place on the E.coli chromosome:
Location for IS
The F integration site determines the order of gene transfer in Hfrs
What is the difference between an Hfr strain and an F+
Possibility of recombination from conjugation
Difference in crossing an Hfr with F- opposed to crossing an F+ with an F-?
Some transferred DNA may be lost following Hfr x FF+ to F- cross cannot lose DNA because the F plasmid is circular, which is retained in the recipient
Hfr cross outcome is with linear DNA that may not be maintained in a bacterial cell
There are multiple insertion sequences (IS) in the E. coli genome
-Orientation of IS determines the order of gene transfer
Example:
Figure of E. coli chromosome Fplasmid > Hfr > IS
What do we expect if the orientation of the Fplasmid is flipped?
The orientation of the alleles changes in the chromosome
Then, the resulting linear DNA sequence will be the same sequence
FINE-SCALE MAPPING BASED ON RECOMBINATION FREQUENCY
Terminology
Merozygote - partially diploid bacteria
Endogenote - recipient genome
Exogenote - transferred DNA
To map chromosome
1. Ensure that recipients (F-) were myozygotes for all genes being considered
2. Select for presence of last gene being transferred
Once DNA is transgerred, frequency of recombination depends on distance between genes
Transferred DNA:
gal+ arg+ met+
Host chromosome:
galargmetRecipient cell is diploid for three alleles
Integration depends on both transfer and recombination
If gal+ is the last to be transferred (start at met+), select only those exconjugastes that are gal+
Time of entry show the direction of transfer and distance between alleles
Example Hft srain . Therefore if we select cells that are leu+ we know that met+ and arg+ must also have
been transferred although not always integrated
Generation of various recombinants by crossing over in different regions
Higher frequency of a genotype occuring:
Because higher the crossover frequency, larger the distance between alleles
Faulty outlooping produces F- plasmid
a) insertion
b) Integrated F factor on Hfr chromosome
c) Excision
d) F`lac with lac+ F plamid now containse chromosomal DNA -> lac+ region
e) Meterozygote
F`lac+/ lacFor a certain period of time, can exist with diploid for two lac alleles
Dominant or recessive? -> relationships can be observed
We expect to be better at transferring lac+ allele/converting genotype
When F` plasmids acquire drug resistance genes usually called R plasmids
Multiple drug resistances are common
Cross-species conjugation
Medical challenge - rapid transmittance
Starting points for creating may gene cloning factors
R plasmids have importance in cloning genes, and practical real world research applications
A plasmid with segments from many former bacterial hosts
Lactococcus lactis used in dairy products
Plasmid contains genes that are acquired from many other bacterial species
Many chromosomal alleles which are scattered throughout the plasmid shows many different integrations
Could be a cause for concern if its antibiotic factors get transferred to bacterial DNA for pathogens
Transformation:
Acquiring DNA fragments from the environment through the bacterial wall
Uptake system
Convert DNA as it is imported into cell
Takes up single stranded DNA to transform the genotype
Artificially treated cells with calcium chloride or electroporation treatment enables cells to take up more
DNA for recombination/transformation
Transformation and linkage
-only small DNA fragments are taken up naturally
Therfore, only genes close together on chromosomal DNA will be co-transformed
We can use this to tell us about distance on the chromosome
Determine order of the three alleles on the DNA
If the two alleles get transferred together frequenty, they are closer together on the chromosome
Method: select for one marker and determine the frequency of the other two markers
Results:
Experiment
Selected Marker
Unselected Marker
1
met+
20% gal+, 2% leu+
2
leu+
8% gal+, 2% met
tells us that met and gal are close together; leu is closer to gal than to met
met
gal
leu
Bacteria exchange by several processes
Transformation
Conjugation
Transduction
Bacteriophages = viruses that attack bacteria
Nucleic acid + protein coat
-Phage T4 parts can spontaneously assemble individually
-Free phage contains chromosomal DNA in its "hat" of protein coat
-Phage infecting a host injects its genomic DNA into cell
Bacteria exchange DNA by several processes
Conjugation
Transformation
Transduction
Transduction
The lytic cycle
-phage T4 infects bacterial cell, injects information (genomic DNA), degrades host chromosome.
Assemble new phages, lysis of host cell
-amplification of phages
Plaques - are clear areas where all bacterial cells have been lysed by phage
Eg) E. coli to low phage concentration to score phage genotypes on a bacterial cell
We can map phage genes on phage "chromosome" because:
1. Several phages with different genotypes can infect the same bacterial cell simultaneously
2. Different phage genomes undergo recombination
(ie) can h+r+ or h-r- phage be produced?
Phage 1
Phage 2
h- r+
h+ rh-rh+r+
yes
Doubling infecting a host with phages of different genotypes to produce recombinants
P
R
h-r+ and h+rh-r- and h+r+
h- = infects two E.coli strains (dark plaque)
h+ = infect one strain (cloudy plaque)
r+ = normal cell lysis rate
r- fast cell lysis rate
Recombination Frequency (RF) = recombinants / total phages
RF = h+r+ h-rAll genotypes
Limitation: cannot detect recombination between phages with the same genotypes
Advantage: We can screen large numbers of plaques and detect rare recombinants as those due to
crossovers swapping portions of a single gene
Transduction: phage-mediated transfer of bacterial chromosomal genes
Two mechanisms (Generalized Transduction)
1. Faulty head transduction
-bacterial genomic DNA instead of phage DNA
2. Specialized Transduction
-prerequesite is integration of
Generalized transduction by random incorporation of bacterial DNA into phage heads
-no lysis
-allows recombination
A phage particle erroneously packages a host chromosomal DNA fragment
-Co-transduction of bacterial genes indicates close linkage
-Mapping is similar to transformation analysis
-Size limitation of DNA package into phage particles
Example: Three genes closely linked: leu+ arg+ lac+
What order are they in?
Infect strain 1: leu+ arg+ lac+ with lytic phage
Isolate phage progeny on a infect strain 2 leu- arg- lacResults:
1
2
3
Gene Marker
arg+
leu+
lac+
60% lac+ and 0% leu+
8% lac+ and 0% arg+
59% arg+ and 8% leu+
arg and leu are never packaged together because they are too far apart
--l--------l-----------------l--arg
lac
leu
Specialized Transduction
Phage transfer bacterial genes through lysogeny-lytic cycle
Lambda Phage (  )
Can integrate its genome into the E.coli chromosome and remain dormant = lysogeny - this
depends on bacterial cell resources
or
Can actively replicate and cause cell lysis = lytic cycle
Production of lysogen
-targetted to a fixed site in the bacterial genome
Single site of integration (*key difference)
-phage factors are required for recombination
-transfer of  prophage during conjugation can trigger lysis
Hfr () x F- lysis [zygotic induction]
*Only genes transferred before  entry will recombine and can be detected
-transfer of  prophage during conjugation will not trigger lysis of lysogenic FHfr () x F- () no lysis
Cytoplasmic factors repress incoming  phage from inducing lysis
-faulty outlooping produces  phage containing bacterial DNA
production or initial lysate
 dgal cn infect but not integrate or recombine: needs help
d = defective  particle
Lambda phage
Can cause E. coli cell lysis
Integrates at a single site in the E.coli bacterial genome
Produces protein factors
Circular genomic structure
Can only be lytic or lysogenic at a time
Production of initial lysate
(i)
Normal outlooping
we expect to see a mixture of  particles
(ii)
Rare abnormal outlooping
dgal can infect but not integrate or recombine
needs help
Transduction by initial lysate
Integration - helper phage provides necessary factors
Recombination - homologous bacterial sequences allow double crossovers
Transduction by initial lysate
Example: gal-/gal+
(i)
lysogenic transductions
dgal is defective
 helper is wild type (has components to integrate)
consequence: can integrate the gal+ allele but does not have to replace the gal- so a stable
diplozygote is produced
(ii)
transductants produced by remcombination
the consequence: replace gal- with gal+
haploid result
DOMINANCE
Complete dominance 3:1 phenotypic ratio
3 dominant : 1 recessive
Incomplete Dominance 1:2:1 phenotypic ratio
Heterozygote is different from parents
Is a combination of characteristics together
eg) Flower colour: Red+White parents = pink progeny
Codominance
1:2:1 phenotypic ratio
Heterozygote is different from parents
Shows characteristics of both parents
eg) calico cats
eg) A-B blood group
Blood Groups
-categorized by how blood reacts with specific antibodies
IA reacts with A antibody
IB reacts with B antibody
i - null allele (no antigen - recessive allele)
called O type blood group - universal donor
-heterozygote:
IA IB - universal acceptor - both kinds of antigens
IA IB blood group is an example of codominance
Method of scoring the phenotype may result in assignment of different dominant/recessive relationships
for the same allele
Eg) Sickle Cell Anemea is caused by an Hb mutation (allele HbS) in hemoglobin A
HbA HbA - wild type or normal; blood cells round; no anemia
HbS HbS - blood cells sickle shaped; anemic
Heterozygote:
HbA HbS - blood cells normal except they are sickle shaped
No anemia at sea level, therefore the wild type is dominant
At higher elevation with low O2 the mutant allele is dominant= anemia
Considering behaviour of allele products (A and S), heterozygous show characteristics of both
homozygous
individuals, HbA is codominant to HbS
Lethal Alleles
If a situation occurs where one of the homozygous classes will not survive, the phenotypic ratio changes
from a 1:2:1 to a 2:1 because lethal alleles cause death of the individual
Eg) yellow mice
Eg) Tailess cats show heterozygousy for a lethal allele
MM - wild type
MLM - tailess
Because the lethal allele is crucial for spinal cord development
MLML - lethal
Kitten dies due to spinal column not developing properly
Therefore there will be a 2:1 phenotypic ratio in Manx cats for tailess individuals
Archibald Garrod, discoverer of inborn errors of metabolism - connected a genetic defect to a defect in
chemical reactions in the body
Genes can control these chemical reactions
George Beadle and Edward Tatum: one-gene-one-enzyme (polypeptide) model
Now we know that multiple genes can contribute to one enzyme.
Beadle and Tatum
Isolated arginine auxotrophs in Neurospora
Auxotroph - strain able to grow only on supplemented media
Organisms unable to synthesize own compounds
Prototroph (wild type) - able to grow on minimal media
Minimal media is water, essential elements, and carbon-sugar source
Able to synthesize biological compounds from the minimal media
-Suggested that all biosythetic pathways consist of a series of steps
-Genes control each step
-Beatle and Tatum used this concept to look at intermediaries from each step in the production of argenine
-They saw that each mutants could survive with some or all of the supplemented compounds (ornithine
and citrulline can "rescue" the mutant)
Supplements: ornithine, citrulline, and arginine
if there is a mutation in arg 1 then it can be helped only by all three
------- ornithine---------------- citrulline ---------- arginine
arg1
arg2
arg3
CHAPTER 6 - Gene Interaction
questions: 4-8, 10-14, 21, 27, 28, 43, 49, 50, 57, 64
Definitions:
pleiotrophic mutation: a mutation that affects several different phenotypic characteristics
haplosufficiency: a single dose of the wild type allele can confer the wild type phenotype in a diploid
cell/organism
haploinsufficiency: a single dose of the wild type allele cannot confer the wild type phenotype in a diploid
cell/organism
dominant negative: a mutant allele that in one dose, as in a heterozygous individual, can eliminate the
wild type function resulting in the mutant phentoype
codominance: alleles which, when combined in the heterozygote show aspects of both homozygotes
incomplete dominance: alleles which, when combined in the heterozygote show a phenotype
intermediate between the homozygotes
prototroph: a strain of organism that can grow on minimal media (media with water, basic elements and a
Carbon-source)
auxotroph; a strain of organism that cannot grow on minimal media but requires nutritional
supplementation
complementation: the production of a wild type phenotype when two mutations are combined in a
diploid
heterokaryon: cell with nuclei of two different types in a common cytoplasm
epistasis: the phenotype of a mutant allele at one gene overrides the phenotype of a mutant allele at
another gene, such that the double mutant has the same phenotype as the first mutant
suppression: a mutation in one gene can cancel the effect of a mutation in a second gene, resulting in a
wild type phenotype
penetrance: the proportion of individuals with a specific genotype who show that genotype phenotypically
expressivity: the degree to which a particular genotype is expressed in the phenotype
Chapter 6 - Gene Interaction
Situations where more than one gene is involved in a pathway
Mendel's findings:
one gene, two alleles
each gene controls a single different character
one allele is completely dominant to the other
ratios:
3:1 phenotype = 1:2:1 genotype (monyhybrid selfed)
9:3:3:1 (dihybrid selfed)
1:1(monyhybrid test-crossed)
1:1:1:1(dihybrid test-crossed)
IF genes in dihybrid are linked, frequencies shift towards parentals
Complexity arises:
-a gene may have multiple alleles
-a gene may be pleiotropic (where a single gene controls more than one character)
-several genes may control a single characteristic
6.1 Interactions between alleles of a single gene: variations on dominance
So far, we have considered only cases where one allele is fully (completely) dominant over another
i.e. phenotype of Aa = AA
monohybrid selfed: 1:2:1 genotypic ratio becomes 3:1 phenotypic ratio
Explanations for complete dominance
Case I: wild type allele is fully dominant to the mutant allele
One dose of the wild type allele is enough for wild type phenotype = haplosufficient
(Figure 6-2) wild type allele is fully dominant to the mutant allele
Example: Haplosufficient
AA versus Aa will give the same phenotype
Case II: mutant allele is fully dominant to the wild type allele
explanation i:
One dose of wild type allele is not enough for wild type phenotype=haploinsufficient
explanation ii:
The mutant allele (dominant negative) makes the wild type allele non-functional
e.g. wild type allele makes 30 units of a protein, and 40 are required for wild type phenotype. A null
mutation (0 units) will be dominant
BB = No resulting functional gene product from B
B+ = B makes none; resulting 30 units
++ = 60 functional
b+=++
b makes 15 units plus 30 units from + makes 45
e.g. gene product forms a homodimer, and one defective polypeptide interferes with its function
brittle bone disease - collagen is a trimer
B+ = BB
B makes + nonfunctional
Two models for dominance of a mutation (figure 6-3)
Incomplete dominance (Figure 6-4)
In four-o'clocks true-breeding red (c+c+) x true-breeding white (cc) give 1 red (c+c+) : 2 pink (c+c) : 1
white (cc)
Heterozygous individual has a phenotype intermediate between phenotype of two homozygous
individuals
Codominant allele: the heterozygous individual has a phenotype showing characteristics of both
homozygous individuals
example: blood groups
Blood groups are categorized by how blood reacts to specific antibodies
each antibody reacts with a antigen on the cell surface
one class of antigen (complex sugar molecules) is controlled by gene I
3 alleles (IA, IB, i) that produce different antigens recognized by different antibodies
IA reacts with A antibody
IB reacts with B antibody
i - null allele (no antigen) reacts with no antibody
Homozygous genotypes and blood group designations:
IA IA - A blood group; IB IB - B blood group; i i - O blood group
Heterozygous combinations
IA i - A blood group; IB i - B blood group; IA IB - AB blood group
IA and IB are dominant to i
IA and IB are codominant: the heterozygote has a phenotype showing characteristics of both
homozygous individuals
Method of scoring the phenotype may result in assignment of different dominant/recessive relationships
e.g sickle cell anemia (Figure 6.5)
Sickle cell anemea caused by a Hb mutation (allele HbS) in hemoglobin A
HbA HbA -wild type, blood cells round, NO anemia
HbS HbS - blood cells sickle shaped, ANEMIC
HbA HbS - blood cells normal and some intermediate shaped cells, except sickle-shaped under low O2,
NO anemia
i) considering anemia HbA > HbS
ii) Considering cell shape, HbA shows incomplete dominance to HbS (intermediate phenotype of some
cells) under normal oxygen, but HbS > HbA under low oxygen
iii) Considering behaviour (migration in an electrophoretic gel) of allele products (A and S),
heterozygous show characteristics of both homozygous individuals, HbA is codominant to HbS
(Figure
6-6)
Lethal alleles - causes death of the individual
e.g yellow coat in mice is a recessive lethal allele
yellow x yellow results in 1 black : 2 yellow (Figure 6-8)
Tailless, a recessive lethal allele in cats (Figure 6-9)
ML/M = tailless
ML/ML - lethal
Archibald Garrod, discoverer of inborn errors of metabolism - connected a genetic defect to a defect in
chemical reactions in the body (Figure 6-10)
George Beadle and Edward Tatum (Nobel Prize, 1958): one-gene-one-enzyme (polypeptide) model
Beadle and Tatum isolated arginine auxotrophs in Neurospora ((Figure 6-11)
auxotroph - strain able to grow only on supplemented media
prototroph (wild type) -able to grow on minimal media
1. mutagenized spores transferred to complete medium
2. asexual spores transferred to minimal media
3. auxotrophs cannot grow, tested for growth on various media
Isolated auxotrophs for a variety of compounds
all segregated as a single gene mutation (1:1 following cross to wild type)
3 arginine auxotrophs mapped to 3 chromosomes - 3 different genes
Beadle and Tatum hypothesized that biosynthetic pathways = series of steps
Arginine has 2 analogs citrulline and ornithine Do these represent different products of steps in the pathway?
Do the mutations represent defects to the different steps? (Figure 6-12)
Hypothesis: If the auxotrophs represent defects to different steps in the pathway, they should be rescued
by a unique set of compounds (Table 6-1)
compounds tested = arginine, ornithine and citrulline
Based on data:
1. Mutations in different genes (auxotrophs) represent unique defects within the biosynthetic
pathway
2. Steps within biosynthetic pathway are controlled by enzymes
3. Each mutation is defective in a different step
(enzyme)
4. Each wild type gene controls a step (makes an enzyme)
Biosynetic pathway consists of a series of steps, each step controlled by a unique enzyme
that is made by a unique gene
One-gene-one-enzyme hypothesis
modified to one gene produces one polypeptide since some enzymes are multiple polypeptides
6.3 Inferring Gene Interactions
The Complementation test
Harebell plant, wild type flowers are blue, 3 plants are found, all with white flowers - mutations in 3 genes,
3 alleles of one gene?
Consider the situation where 2 lines ($ and £) are alleles of one gene (w1), the third line (¥) is an allele of a
second gene (w2) Figure 6-15
If we make pairwise crosses amongst the three lines ($ x £, $ x ¥, £ x ¥) progeny of crosses between
alleles of the same gene will show no complementation, progeny of crosses between alleles of
different genes will show complementation (Figure 6-15)
In the first cross, the F1 progeny have no wild type copy of w1 gene
In the second cross, the F1 progeny have one wild type copy of both the w1 and w2 genes (Figure 6-15)
RULES FOR A COMPLEMENTATION TEST
1. can only be done with recessive mutations
2. If the mutations are in different genes, the two mutations will complement one another
(progeny will be wild type)
3. If the mutations are alleles of the same gene, the two mutations will not complement one
another (progeny will be mutant)
Heterokaryon = variation on a complementation test
-cells have two nuclei in a common cytoplasm
-products of two nuclear genomes act together in the common cytoplasm (Figure 6-16)
Using the complementation test to find a novel gene:
A plant mutant for FORKED1 (FKD1) has an open vein pattern
We mapped FKD1: 2 recombinations out of 3000 plants to a CAPS marker on chromsome
3
(0.067 cM), meaning that FKD1 must be within a region of about 40 genes
One line, with a mutation in gene At3g63300, has a similar phenotype to fkd1-1
Crossing to this insertion line produces mutant seedlings (i.e. failed to complement),
therefore these are alleles of the same gene and fkd1-1 must be an allele of At3g63300
Transformation of fkd1 mutant with the cloned FKD1 gene results in a wildtype phenotype
= molecular complementation
Clicker question 1:
The gene wg controls formation of wings in Drosophila. For a cell to become part of the wing, it requires 50
units of the wg gene product; if a cell has less than 50 units, it will not become part of the wing. The wild
type wg+ allele makes 30 units of the wg gene product; a mutant allele (wg) makes 0 units of the wg gene
product. Which of the following statements are true:
A) the gene is haplosufficient
B) the mutant allele is dominant
C) the wild type allele is dominant
D) all of the above
E) none of the above
Clicker question 2:
Nasturtium flowers are normally red, but orange, white and yellow varieties exist. You make crosses
amongst different plants with the following results:
cross
progeny
RxR
all red
WxW
all white
YxY
all yellow
OxO
1 red : 2 orange : 1 yellow
RxY
all orange
OxY
1 orange : 1 yellow
OxW
1 red : 1 yellow
Which of the following is consistent with the above data?
A) R and Y result from incompletely dominant alleles
B) orange individuals are heterozygous
C) the allele causing red is dominant to the allele causing white
D) all of the above are consistent
E) none of the above are consistent
Mutants
Harabells are usually blue. Two white mutants are crossed to give all white progeny.
a) the mutants carry alleles of the same gene
Chapter 3 - 9:3:3:1 independently assorting genes
Chapter Four - shift towards parental gametes linkage not equal to 9331
Chapter Six - 9:3:3:1 genes interlinked ; where in same pathway the phenotype is indistinguishable
meaning 0:9:9:3:4 ratio
Analysing Double Mutants: variations on the 9:3:3:1 ratio
1. no gene interaction is indicated by 9:3:3:1 ratio
eg) corn snakes with skin colour due to two unlinked genes o and b.
o+ = presence of orange pigment
o = absense of pigment
b+ = presence of black pigment
b = absense of black pigment
o+/- ; +/o+/- ; b/b
o/o ; b+/o/o ; b/b
= camouflaged
= orange
= black
= albino
The 9:3:3:1 ratio indicates that the genes act in independent or separate pathways (although they are
independently assorting)
Colourless precursor plus b+ = black
Colourless precursor plus o+ = orange
Both o+ and b+ present = both expressed in camouflaged
Neither o+ or b+ present = neither expressed = albino
2. The 9:7 phenotypic ratio
9:7 ratio indicates one of several explanations:
- complementary gene action
- same pathway (no distinguishable intermediary inter-individual)
- one gene regulates the other
3. Genes in the same pathway may show an epistatic interaction
blue-eyed Mary - wild type = blue
two mutants: white (w/w) and magenta (m/m)
Cross: w/w m + /m + x
w+/w+ m/m
White
magenta
F1:
w+/w ; m+/m
Blue
F2:
9 w+_ m+_ (blue)
3 (magenta)
3 (white)
1 w/w m/m (white)
Two possible pathways:
w+
m+
white ------------ > magenta ---------- > blue
m+
w+
magenta -------- > white --------------- > blue
The first pathway is correct because the upstream gene covers up the downstream gene. Once w is
defective, the second gene still does not matter and the individual is still white. We see in the progeny a 9
blue : 3 magenta : 4 white
Epistasis = w is epistatic to m; w covers up m phenotype; w is upstream of the m in the gene pathway;
progeny result in a 9:3:4 ratio where 3 is recessive
12:3:1 is dominant epistasis
Labrador Retrievers
Golden, Chocolate, and Black dogs
Yellow gene (Y) required for deposition of pigment in fur - epistatic to the alleles B (black) and b (brow)
Golden Labs are homozygous for e/e but you cannot tell if they have black or brown alleles because the
pigment is not deposited in the fur
Black
B/B E/_
Chocolate
b/b E/_
Golden
_ /_ e/e
Coat colour also controlled by B gene
Siamese Cat - temperature sensitive allele of B
B gene pigment is functional is cooler regions (face ears extremities tails paws)
B is not expressed along body where the cat is white/beige
Genes controlling spore colour in Neurospora
Al+ = orange
Al = albino
Ylo+ = orange
Ylo = yellow
Create a diploid meiocyte by crossing the haploid parents
Spore progeny created from the meiocyte are 1:2:1
¼ al+ylo+
orange
¼ al+ ylo
yellow
¼ al ylo+
albino
¼ al ylo
albino
Suggests that al is upstream of ylo
4. In foxglove flowers
D= dark red, d = pink; w allows pigment throughout petal, W pigment is only in the throat of the flower; W
is epistatic to alleles of D
Dominant epistasis results in a 12:3:1 ratio
W is controlling D
12 White - W_ ___
3
1
Epistatic allele is upstream in the pathway so that is does not matter what the downstream allele is.
5. Suppression
A mutation in one gene reverses the affect of a mutation in another gene, resulting in a wild type phenotype
in the double mutant
10
3
3
10
6
13
3
Eg) purple eye colour in Drosophila and suppressor su
Parents:
p/d ; su+/su+
Purple
x
pd+/pd+ ; su/su
Red
F1 red
F2 9 red: 3 red : 3 purple : 1 red
Explanation:
- one bypasses the other
- suppressor modifies a binding partner
Modified 9:3:3:1 ratios provide important information about the mechanisms of gene action within
pathways
PENETRANCE
The percentage of individuals with a given genotype who exhibit the phenotype associated with that
genotype is penetrance. Penetrance is expression within a population whereas expressivity is in the
individual.
Example
B _ = normal vision
bb = blind
But of the 100 people of genotype bb, only 20 are blind, therefore allele b has the penetrance of
20% this means they have a tendency to being blind
EXPRESSIVITY - the extent tow hich a given phenotype is expressed in an individual
Example
A _ = normal pigmentation
aa = albino
variable penetrance
variable expressivity
variable penetrance and expressivity
Chapter 7: DNA Structure and Replication
Introduce molecular biology/genetics
Determination of genotypes at molecular level:
Gene expression Gene
replication
Distinguishing classical Mendelian Genetics from modern Molec. Genetics
What is genetic material?
Main cellular components:
Polysaccharides (sugars), lipids, proteins, DNA, RNA
Pre-1944 it was believed that protein was more likely than DNA to be the genetic material because:
- 20 standard amino acids known, only four bases in DNA, so proteins could house much
information
- DNA sequence thought to be repetitive
(if same four bases repeated in same order so then little info content)
Frederick Griffith 1928
Transforming Streptococcus R cells into S cells
-His experiment showed that a subcomponent of a cell could house genetic information. Isolated two
different strains of bacterial
-S (wild = smooth; bacteria coating; when infect mice will kill mouse)
-R (mutant = rough; bacteria is incapable of producing a polysacharide coat; unable to infect mice)
a)
b)
c)
d)
S strain live cells mouse dies
R strain mouse lives
S strain heat-killed cells mouse lives
S strain heat-killed and R strain mouse dies
Conclusion:
A component from dead S cells is taken up and transforms R cells with virulence properties
The transformed R cells divide and propagate as S cells
- hereditable material has been acquired
R cells have acquired the genes for virulence.
Avery MacLeod and McCarthy 1944
Which specific component of heat-killed S cells is required for transformation?
Extract components from the S strain
If enzymes were used to destroy polysaccharids, lipids, RNA, and proteins, the transforming ability was not
affected and the bacteria still was able to kill the mice.
In the case of destroying the DNA component of the S strain, the mouse lived; therefore the DNA contained
the genetic material that was transformed
Hershey and Chase 1952
Differential labeling of phage components:
-proteins 35S-labeled
-DNA 32P-labeled
Is it the DNA or the protein coat that is responsible for transformation of genetic information?
-proteins 35S-labeled
allow radioactive phage to the E. coli cell
then separate protein coats from E. coli
Most of radioactivity recovered in phage ghosts (away from E. coli cell)
-DNA 32P-labeled
DNA is the radioactive component
When phage particles attach to the cell, inject DNA
Most of radioactivity recovered in E. coli cell
Conclusion: The majority of radioactivity injected by phage into E.coli was 32P-labeled DNA. Phage DNA
drives the production of new phage particles
1) What are the chemical and physical properties of DNA?
2) What is the 3-d structure of DNA?
Requirements to store genetic information:
1. Stability
-must maintain structure during the lifetime of an organism
-most organisms use DNA; some viruses use RNA
2. Replication
-accuracy and efficiency
3. Informational complexity
-diversity of life forms
-able to specify why each life form is different
4. Mutational capacity
-balance between replication accurancy and ability to change
-crucial to evolution
Structure of the four DNA nucleotides
-nitrogenous base
difference between nucleotides
-deoxyribose sugar
-phosphate
only one phosphate group individually
two additional phosphates are dAMPs which are builing blocks of structure
Purine nucleotides
A- adenine nitrogenous base
G- guanine nitrogenous base
Pyrimidine nucleotides
C- cytosine nitrogenous base Tthymine nitrogenous base
Originally, biologists thought that DNA was too simple to house genetic information
Chargaff's DNA base composition results from different organisms
Accurately determining molar composition
Different bases can be present in different relative ratios
Shows that base content is dissimilar in different organisms
Therefore, different species can have different genomes
Rules:
1. Total base ratios A's + G's = C' + T's
2.A's=T's
G's=C's
This tells us that the purine ratio equals the pyrimidine bases
Consistent patterns
What is the 3-D structure of DNA?
James Watson and Francis Crick, 1953
The era of molecular biology
Watson and Crick deduced DNA structure from the information of earlier times
- Chargaff's Rules
- X-ray diffraction of DNA fibres
- Molecular Modelling
Rosalind Franklin's X-ray Diffraciton Data
- repetitive structure: uniform geometry
- Helical structure
- Double helical: two strands
Linus Pauling
- working out protein structure
-
alpha helices produce a similar x-ray diffraction pattern to DNA
therefore, DNA must be helical as well; but with two strands
The structure of DNA
-Double helix
two strands in winding helix
-Bases in the interior
nitrogenous bases play critical role in holding strands together
-Sugar-phosphate backbone on exterior
exposed to aqueous solvent
The Chemical Structure of DNA
Strand polarity
- 5` to 3` end
-antiparrallel arrangement means the other strand has opposite orientation
Hydrogen bonds between bases
- G-C (3)
- A-T (2)
- stability
maintains overall parameters/distance
interactions between nitrogenous bases: hydrogen bonds hold together
Phosphodiester bond links nucleotides
- gives the strands their orientation
- 5` and 3` ends have no more nucleotides attached
-opposite orientation of two strands:
antiparrellel arrangement is the only way to enable H-bond interaction
Nucleic Acid:
Formed by joining nucleotide units
nucleotide units are joined by the phosphate groups
Chapter 7 Practice Problems: 1, 2, 11, 16, 19, 24, 28
Representation of the DNA double helix
-stacked and planar base-pairs
-Space-filling representation: each atom as a sphere:
shows 3-D orientation
major groove: large space where no atoms as present as helix winds
minot groove: also wrapping around length of the helix
Grooves cause nitrogenous bases to be exposed
chemicals and proteins can interact with DNA nitrogenous bases in grooves
-Uniform geometry and helical width explain Chargaff's data
The only way to accommodate the distance between phosphate backbone/ two DNA strands is if a purine is
paired with a pyramidine
This is the helical width (20A)
The distance between adjacent bases is 3.4A
The distance per revolution of the helic is 34 A
Therefore, there are 10 base pairs per revolution of DNA double helix
Question: Do the base-pairing rules observed within the double helical structure also control the process of
DNA replication? Or is replication like a photocopying process?
Alternative models for DNA replication
Semiconservative replication
Two blue strands
One parental + one new strand (1 blue 1 yellow)
one parental + one new strand (1 blue 1 yellow)
Conservative replication
Two parental strands (two blue)
Two parental strands Two new daughter strands (two yellow)
Dispersive replication
Each strand is chimeric of
Two parental strands ~50%old and ~50% new
Patchwork of blue/yellow
Meselson-Stahl Experiment
Strategy: Distinguish newly synthesized strands from parental strands using different isotope labeling
(different weight)
Separate DNAs with different relative densities
-labelled parental DNA by growing E.coli in 15N medium for many generations (heavy DNA)
-transferred cells to 14N (which makes light DNA in replication)
-extracted DNA after 1 and 2 generations (two rounds of DNA replication)
-centrifuge in CsCl gradient
-separate DNA of different densities
-heavy nitrogen DNA will move more quickly to the bottom of tube
Predictions of replication models
Conservative Replication
Would see first generation to have half heavy and half light DNA
Second generation will see more light DNA to one heavy DNA
Blue heavy DNA
Yellow light DNA
Parental
Blue
Blue
Yellow
Yellow light DNA
Yellow light DNA
Semiconservative Replication Predictions
Would see first generation at middle of tube
Second generation will have half in middle, half light DNA rising to top
half Blue/Yellow
Parental
half Blue/Yellow
Yellow light DNA
Heavy Blue
half Blue/Yellow
Yellow light DNA
half Blue/Yellow
Dispersive Replication Predictions
Would expect that after successive generations, the DNA becomes lighter and lighter when cultured
in light nitrogen.
¼ Blue / ¾ Yellow
Parental
half Blue/Yellow
¼ Blue / ¾ Yellow
Heavy Blue
half Blue/Yellow
¼ Blue / ¾ Yellow
¼ Blue / ¾ Yellow
RESULTS
The cesium chloride centrifuging showed the semiconservative replication model
Semiconservative DNA Replication
Base-pairing to a template strand determines what is next added to a synthesized strand
Both parental strands act as templates
Strands at a replication fork
Deoxyribo sugar and phosphate
Reaction catalyzed by DNA polymerase
Uses DNTP
Hydroxyl group attacks phosphate
Phosphodiester bond joins units
*important consequence for replication
Property #1
DNA synthesis always occurs 5` to 3` end
But: As a replication fork proceeds in one direction along the DNA template
Both strands are replicated simultaneously
DNA replication at the growing fork
Directional 5` to 3` synthesis of both strands
Fork movement causes leading and lagging strands
Leading strand: strand in which DNA synthesis proceeds in same direction as replication fork moves
Lagging strand - DNA synthesis proceeds in opposite direction to replication fork movement
Property #2
DNA polymerase requires a primer (short nucleic acid)
Primase enzyme synthesizes an RNA primer
8-12 nucleotides long
DNA polymerase adds first dNMP onto 3` end
Synthesizing the lagging strand
1. primase synthesizes short RNA oligonucleotides (primer) copied from DNA
2. DNA polymerase iii elongates RNA primers with new DNA
Okasaki fragment: has two components: RNA primer and ..
ICLICKER example
DNA sequence of two strands of one DNA molecule:
5` AGGGCTAAG 3`
3` TCCCGATTC 5`
DNA pol I works as a 5` to 3` exonuclease
DNA Polymerase
First DNA polymerase activity characterized was E.coli DNA pol I
- by Arthur Kornberg
- not processive:
only short DNAs synthesized; could not synthesize a long leading strand
Later DNA polymerase III was discovered
- processive: can synthesize long strands of DNA
- replicates most of the chromosome
Sythesizing the lagging strand
1. Primase synthesizes short RNA oligonucleotides (primer) copied from DNA
2. DNA pol III elongates RNA primers with new DNA
3. DNA pol I removes TNA at 5` end of neighbouring fragment and fills gap
4. DNA ligase connects adjacent fragments
a. Ligation: taking two DNA molecule fragments and fusing to make one
DNA polymerase I has a 5' to 3' exonuclease activity. This activity removes the RNA portion from the 5' end
of an Okazaki fragment while the polymerase activity replaces the degraded RNA sequence with new
DNA
sequence - notice that it uses the adjacent 3' end of the neighboring Okazaki fragment as primer (DNA is
not added to the Okazaki fragment whose RNA portion is being degraded)
-DNA synthesis is continuous on the leading strand
1 RNA primer and synthesis to end of molecule
-DNA synthesis discontinuous on the lagging strand
Many Okazaki fragments and primers stitching together of many fragments.
DNA ligase joins Okazaki fragments
Creates new phosphodiester bonds
This enzymes is very particular for the substrates it uses
5` phosphate and 3` hydroxyl required for joining to occur
E. coli chromosome replication visualized using [3H] thymidine
Chromosom after one round of replication
Autoradiograph diagram Interpreation of Semi Conservative
Single strand
double strand of one radioactive DNA strand
of radioactive DNA
and one strand not radioactive
Chromosome during second round of replication
Autoradiograph
diagram Interpreation of Semi Conservative
Two strands replicated simultaneously
Two radioactive strands
This tells us both strands of DNA used in replication
Therefore, a fork occurs because DNA strands
separate
.
Step 1 - label synthesized DNA with low levels of radioactivity (early pulse)
Step 2 - next label with higher radioactivity amount (late pulse)
E. coli circular genome has a single replication start point
- 1 origin of replication
- 2 replication forks Bidirectional replication
Bidirectional replication in eukaryotes
1. Pulse with high [3H] thymidine concentrations - Hot
2. Pulse with lower [3H] thymidine concentrations - Warm
Replication of chromosomes in Drosophila cells. Only a very small region of a chromosome is illustrated
here.
Such studies have revealed clusters of active replicons
Replicon - unit length of DNA that is replicated starting from a single origin of replication.
Chromosomes in eukaryotic organisms such as humans may have thousands of origins of
replication.
Replication of linear eukaryotic chromosomes
-Replication forks from adjacent replicons will eventually meet and the synthesized DNA strands
will be ligated.
-Replication time for the entire chromosome is greatly reduced by starting at multiple origins.
..
SUMMARY
E. coli (and other prokaryotes) use a single origin of replication
- bidirectional replication - new replication initiation may occur before the
previous round of replication is complete (faster cell growth)
Eukaryotes use many origins - bidirectional replication
- one initiation event at each origin; new initiation may occur before the previous.
Opening the helix at the origin in prokaryotes
Steps:
1. Proteins specifically recognize sequences (box elements) unique to the origin
2. DNA is melted (strand separation) 3. Proteins keep the strands separated
Opening the helix at the origin
Origin = recognition sequences (box elements) + AT richregion
•DnaA - protein that recognizes box elements - makes sequence-specific interactions
•DnaB - protein (helicase) that recognizes DnaA - further helix opening during DNA synthesis
•Single-stranded DNA binding proteins - keep the strands apart
1. DnaAbindsboxes
2. Co-operativefurtherbinding of DnaA coats the origin and leads to melting in the AT-rich region
3. DnaB helicase recognizes DnaA
- displacement of DnaA
4. Binding of replication machinery - replication forks are now established
Opening the helix at the origin in eukaryotes Yeast model
• Origin recognition complex (ORC) - multi-subunit
- binds the lone DNA box element
• Cdc6 and Cdt1 - bind ORC and regulate initiation
• Helicase - activity is inhibited by Cdc6 and Cdt1 - strands will not be unwound and separated
and DNA polymerase will not bind without functional helicase activity
DNA replication in eukaryotes
What is the importance of the regulatory Cdc6 and Cdt1 proteins?
1. DNA replication must only occur during S phase
2. Chromosomes must only be replicated once
2n
4n or n
2n (in haploid fungi)
Additional Chapter 7 Problems: 3, 5, 14, 18, 21
Opening the helix at the origin in prokaryotes
Steps:
1. Proteins specifically recognize sequences (box elements) unique to the origin.
2. DNA is melted (strand separation)
3. Proteins keep the strand separated
Helicase protein at each replication fork
DNA Replication in Eukaryotes
1. DNA replication must only occur during S phase
2. Chromosomes must only be replicated once
2n 4n ONCE!
Additional Chapter 7 Problems: 8, 13, 15, 22, 26
ICLICKER
Which one of the following statements about lagging strand synthesis is incorrect?
d) DNA pol III is used to replace the RNA primer sequence with DNA sequence
Correct statements of lagging strands
a) an RNA primer is used and eventually removed from the 5` end of the Okazaki fragments
b) Laggging strand synthesis requires the production of many Okazaki fragments
c) Lagging strand synthesis occurs in a 5` to 3` direction
d) DNA pol I is used to replace the RNA primer sequence with DNA sequence
e) DNA pol III synthesizes MOST DNA
REPLICATION
Regulatory proteins: Cdc6 and cdt1
There are specific stages in the cell cycle; DNA can only be replicated in S phase and only once so that
only
two copies result
Gap Phase synthesizes regulatory proteins
Pre-replication complex forms:
-Origin Recognition Complex (ORC)
-Cdc6
-cdt1
-helicase (not active for replication in G1, G2 or M, activates in S phase)
G1 Phase
-Cdc6 and Cdt1 inhibit the helicase
Transition between G1 and S phases
-removal of protein components:
-Cdc6 degraded, Cdt1 inhibited
Helicase activated
S phase
-Helicase creates a replication fork
-DNA pol complex can begin recruiting
-only initiate replication once, since regulatory proteins have been removed, helicase can only be activated
at the one previous time
The Replisome
-All the components required for DNA replication
Functions:
1) Progressively unwind the helix and keep strands separated
2) Catalyze DNA synthesis (phosphodiester bond formation) on both leading and lagging strands
-Replisome has both kinds of catalytic activity
-two DNA pol III complexes for each lagging and leading strand
3) Join the Okazaki fragments
-Replisome has activities that can remove RNA primer and join the frags
Polymeraze components
Two associated DNA pol III complexes
Beta clamp for processivity
DNA looping
(portion of DNA is pulled around to be read by DNA pol III
so that DNA pols can move in the same direction, but can still synthesize in a 5` to 3`
direction.)
Primase for lagging strand synthesis
Replisome at work
Unwinding components
Helicase - strand separation
Single-stranded binding proteins
(ensure each individual strand stays apart after helicase separates them)
Topoisomerase
Progressive fork movement and DNA unwinding creates helical strain
Parental Duplex is unwound
-remove helical twist
In circlular DNA, such as in E. coli:
-DNA molecule does not like this, so what happens to compensate is an over-winding of upper DNA
regions = Supercoils
-as replication fork moves around circle, the supercoils would create enough tension to stop
helicase working, therefore topoisomerases are required to unwind the supercoils as they form
Topoisomerases relieve the strain from supercoils
DNA gyrase is a type of topoisomerase
1) DNA gyrase cuts DNA strands
2) DNA can rotate to remove the twists of the coil
3) DNA gyrase rejoins the DNA strands
Replisome at work
Finally, must deal with Okazaki fragments to finish off lagging strand synthesis
DNA pol I and DNA ligase are components of the Replisome as well
TEST QUESTION ON ABOVE MATERIAL
**** what would be the role of this enzyme activity and what is it a component of***
In eukaryotic organisms the chromosomal DNA exist as chromatin
Complex of DNA + histones = nucleosome
Nucleosomes must be displaced and then replaced behind the replication fork on the daughter molecules
There are not enough nucleosomes from the parent molecule for both daughters
Assembling nucleosomes during DNA replication
Chromatin assembly factor
-binds to newly synthesized histones
-assembles nucleosome structure
The replication problem at chromosome ends
Gap cannot be filled because there is no primer
Observation: Eukaryotic chromosomes have repetitive sequences at their ends called telomeres
Telomerase: A protein-RNA complex adds the repetitive sequences to the 3` ends of linear DNA molecules
1. Provides a template (RNA) for DNA synthesis
2. Enzyme activity that uses RNA as template for DNA synthesis (reverse transcriptase)
Lengthening of the 3` Overhang
A region of the telomerase RNA acts as a template
-enzyme binds to ends of linear chromosomes
-base pair interaction between RNA/DNA
-this leaves an overhang region
telomerase enzyme can add on bases
there is a repeated sequence at both ends (eg. AAC ------- AAC)
Reverse transcriptase:
5` to 3` synthesis
Repetitive template sequence allows re-positioning
Replication of complementary strand
A primer is synthesized
Polymerase fills in the gap
The primer is removed and ligase seals the gap
Only a small repeated sequence is lost at the end of the chromosome during this replication.
Chromosome spread
-Telomeres can be recognized by a fluorescently labeled DNA probe complementary to the
repeated
sequence
-eg picture: see four telomeres
DNA replication has occurred
Two telomeres per DNA molecule
-Intact nuclei
Telomerase: Links to human disease and cell aging
Germ cells: lots of telomerase activity
Must maintain chromosomal activity to pass on genetic information to sex cells and gametes to
offspring
Somatic cells: little telomerase activity
-cells enter into a senescence phase
-loss of telomere integrity
Cancer cells: many have over-active telomerase
-immortalization
Deficiencies in telomerase
-Werner Syndrome
-Dyskeratosis congenita
=Premature aging
Chapter 8 RNA: Transcription and Processing
Questions:
How is genetic information encoded in the nucleus (DNA) transmitted to the cytoplasm where proteins are
synthesized?
What molecule is synthesized in the nucleus and then moves into cytoplasm that could carry such genetic
information?
Pulse-Chase Experiment
Pulse label a eukaryotic cell with radioactive uracil
(incorporate radioactivity into a newly synthesized molecule in a cell)
(use uracil because it is in RNA and will give newly synthesized only)
Then chase with excess non-radioactive uracil
(we can follow the molecule as it moves through cell)
The molecule will breakdown
Radioactivity disappears
This experiment tells us where the molecule is synthesized, where it goes, and how long it survives as that
molecule in the cell
Pulse: Shows us where RNA is made
We see radioactivity in the nuclues where the molecule is synthesized
Chase: Eukaryotic RNA moves from nucleus to cytoplasm
The molecule (RNA is "chased" or followed by seeing radioactivity) temporarily accumulates in the
cytoplasm
Let us see the half-life of RNA in the cell
- because can track where radioactive uracil is going without continuing to regenerate new
mRNA without more r.U.
Results
We see some RNA in cytoplasm and some (few) left in nucleus
snRNA in nucleus
tRNA, mRNA, and rRNA in cytoplasm
RNA = ribonucleic acid
RNA is a precursor to protein
-messanger RNA
Carries the "genetic message" from genes in the nucleus to ribosomes in the cytoplasm (eukaryotes)
Properties of RNA
Structural features in RNA are similar and different to DNA
RNA
DNA
Sugar:
Ribose
Deoxyribose
Difference that RNA has a 2` hydroxyl group; DNA lacks that O at 2` C
Nucleotides:
Purines:
Pyrimidines:
Adenosine
Guanine
Cytosine
Uracil
Adenosine
Guanine
Cytosine
Thymine
Summary of Structural Differences
- Contains ribose sugar
o More reactive and less stable due to hydroxyl group
- Contains uracil (U) base instead of thymine base
o U pairs with A
- Can be single-stranded or double-stranded
o Structural complexity
o Internal base pairing within one strand
Classes of RNA
1. Coding RNA
Informational RNA - provide informational template for protein synthesis
a. Messenger RNAs (mRNA)
2. Non-coding RNAs
Functional RNAs - is not translated into a protein
Do not code for protein sequence
Functional/Non-Coding RNAs
- Transfer RNAs (tRNAs) - transport aa to the ribosome
- Ribosomal RNAs (rRNAs) - component of ribosome
- Small nuclear RNAs (snRNA) - RNA processing
Splisosome used to remove introns from RNA species
- Micro RNA (miRNA) - inhibit gene expression
Ability to bind to prevent protein synthesis
Functional RNAs can vary in size/length and therefore overall complexity and roles
Transcription = production of RNA from DNA template
Transcription of DNA translation protein synthesis
Eukaryotic Cell
-mRNA moves from nucleus to cytoplasm
-functional RNAs may reside in any cell compartment
RNA synthesis
Same as DNA synthesis:
-always synthesize 5` to 3`
-complementary basepairing same
-template read 3` to 5`
-NTPs serve as the substrates
-catalytic mechanism of breaking phosphodiester bond similar
Ability to base pair to different nucleotides dictates the sequence of RNA
- Uracil to Adenosine
Comparison: Transcription of RNA from DNA template
Similarity
-5` to 3` synthesis and basepairing rules specify the addition of the next nucleotide
Differences
-RNA polymerase does not require a primer
-Only one strand is synthesized (continuous leading strand) and therefore one template strand only
-for each gene, one DNA strand acts as a template
-other strand = non-template strand
Different strands serve as templates for different genes
Sequences of DNA and transcribed RNA
Non-template strand = Coding strand
RNA is complementary to the template strand, identical to the non-template strand (except U for T)
Coding strands predicts amino acid
-based on the direction of the RNA transcript arising from the gene
we can equate the directionality of the RNA to the directionality of the DNA gene
5` end of the RNA defines 5` end of the gene
RNA transcription "flows" from upstream to downstream
5` to 3` direction
a DNA region is "upstream" of the 5` end
a DNA region is "downstream" of the 3` end
Rule: For most genes there will be a conserved transcription start point and a conserved end point
Sequence signals must define the start and end points
Steps in transcription:
1. Initiation - rate-limiting step, dictates amount of RNA that can be produced
- Start making the RNA chain (slow step)
i. Bind the template
ii. First phosphodiester bonds are formed
2. Elongation
- Continuous synthesis of the RNA chain (faster)
- RNA polymerase movement along the template
3. Termination
- Stop synthesis
i. Phosphodiester bond formation ceases
ii. RNA transcript released
Bacteria have one RNA synthesis complex: the polymerase
-multi-subunit complex
-good model is the RNA pol complex in E. coli
Core enzyme
Contains four different subunits (alpha beta beta` and gamma)
Elongation and termination
Has catalytic activity that can create phosphodiester bonds
Holoenzyme
Same composition of core enzyme plus the sigma factor
Initiation
Recognizes sequence elements to do the initial binding
Which of the following statements about bacterial RNA polymerase is correct?
ANS: It exists in one form called the holoenzyme which contains a
subunit called the sigma factor
How does RNA polymerase know where to start transcription?
Approach: Compare many genes and many start sites. What is in common?
Conserved DNA sequences?
E.coli promoters
Promoter - set if DNA sequences that are required to initiate transcription
- recognized by RNA polymerase
consensus sequence
most common nucleotide among collection, but a gene may have some variation
box elements: a conserved secquence
serve as binding sites for RNA pol complex
RNA pol will start transcription at the +1 position
Summary
E.coli promoter genes contain two minimal elements (core elements)
- the -10 and -35 box
- are located upstream of transcriptional start site
- conserved sequences (consensus)
- conserved distances from start site, which is +1
- strong promoters (high level RNA synthesis) are closer matches to the consensus
Transcription in Prokaryotes
Transcription Initiation
a) RNA pol binding to promoter
-the -35 and -10 region - recognized by sigma factor
b) Initiation
-of phosphodiester bond formation
c) transcription bubble forms
d) RNA synthesis
sigma factor can then dissociate
Transcription Elongation
Transcription buble moves along with the RNA pol complex
Short RNA-DNA hybrid is maintained (10 base-pairs) throughout elongation
Anything that disrupts this will trigger premature termination
As RNA pol moves along, it needs to let DNA rewind after
Unwind with helicase then rewind the DNA double helix
Many RNAs can be simultaneously transcribed from a single gene
-initiation before elongation complete
-looks like christmas tree light structure
-longest RNA transcripts towards end of gene
-we see RNA strands emanating off the gene
- multiple RNA pol on same gene
-example: rRNA transcription in frog oocytes - very active transcription
Question: What defines a transcription stop point?
Answer: Sequence and structure in the RNA just synthesized
Two elements required in RNA transcript:
1) Hairpin structure (stem-loop)
2) Poly U stretch
There is an intrinsic terminator on the RNA transcript
Intrinsic: RNA pol complex on its own without any other RNA factors recognizes this RNA feature
The RNA transcript has ability to base pair with itself, antiparrallel base pair interactions, so are
forming a helical structure in the hairpin stem-loop
At the 3` end of the RNA there needs to be a consequetive stretch of U's
This must be positioned at the base of the stem-loop structure
Termination: intrinsic mechanism
-RNA-DNA hybrid is weak
The base pairing between U's of RNA and the A's of DNA is weak
-Hairpin helps destabilize the hybrid
-RNA then dissociates from DNA template
Transcription in Eukaryotes
-More complex than in prokaryotes
-larger genes and genomes
-more complex gene regulation
-polymerases are larger
-accessory factors
-3 RNA pol (eukaryotes) vs. 1 in prokaryotes
-RNA pol I - ribosomal RNAs
-RNA pol II - mRNAs (and snRNAs)
-RNA pol III - tRNAs and 5SRNA (small RNA component of the ribosome)
Initiation in Eukaryotes
RNA pol complex does not recognize promoters on its own
-unlike the bacterial RNA polymerase holoenzyme
General transcription factors
-several large complexes that recognize promoter elements and/or modify RNA pol activity
-Somewhat analogous to the sigma factor
-Stimulate RNA pol II transcription
-Transcription Factors II (TF IIs)
TFIID, TFIIE, TFIIF etc
Transcription Initiation in Eukaryotes
TFIID - a complex containing several proteins including:
TATA box binding protein
The TATA box - contains the sequence TATA
- correctly positions the RNA pol complex
- more transcription factors bind to form the pre-initiation complex
Pre-initiation complex
-DNA is melted
-RNA pol has bound
-no phosphodiester bond formation yet (not activated yet)
Initiation to Elongation
-RNA polymerase structural change
-RNA pol II begins elongation
1. Phosphorylation of RNA polymerase (by TFIIH)
2. Continuous RNA synthesis begins
3. Transcription factors dissociate from the polymerase
RNA processing - in the eukaryotic nucleus
All eukaryotic RNAs undergo processing events to convert a precursor (primary) transcript into its mature
form
Prokaryotic and eukaryotic transcription and translation
Prokaryote: coupled transcription-translation
Ribosomes can start synthesizing the proteins on one end of mRNA while transcription is still occuring on
that mRNA
Eukaryote: transcription in nucleus
The transcript is converted into a mature form of mRNA
Transportation of mRNA from nucleus to cytoplasm
Translation can then occur because ribosome recognizes mRNA
Transcription Elongation Termination:
Addition of a polymeric (poly) A tail
Endoneclease recongizes a sequence element and then cuts 3` of it
Eukaryotic genes encoding proteins contain intervening sequences
Intervening sequences = introns
-don't usually code for amino acids in proteins
-removed after transcription
Protein-coding portions of genes = Exons
-present in final mature mRNA
Splicing
-Introns are transcribed along with exons in the primary transcript
-Introns are removed and the exons are spliced together
Question: How are the correct sites for splicing chosen?
Conserved sequences specify intron splice sites
GU-AG"rule"
GU at 5` end of pre-mRNA and AG at 3` end
The mechanism of Splicing (write this out a few times/memorize/final)
-moving of phosphodiester bonds
Reactions in exon splicing
First transesterification
Branch point "A" is involved in a 3` to 5` phosphodiester bond
PLUS a 2`-5` phosphodiester bond
To perform this reaction need a free hydroxyl group; the only free one in this case is the 2` hydroxyl from
branch point "A"
There is a transfer of a phosphodiester bond
This can only be done in RNA
Because of the 2` hydroxyl, RNA is more reactive; DNA lacks the hydroxyl
Splicing must be an accurate process
Splicing at same intron position
Complex pattterns of eukaryotic mRNA splicing occur
Highly dependent on spliceosome
During splicing, the mRNA associated with a large RNA-protein complex is called a spliceosome
Eukaryotic mRNA introns are called spliceosomal introns
Roles of the Spliceosome
1) Recognize the conserved intron elements
2) Help fold the intron into the correct 3d shape
- RNA catalysis
3) Regulate the splicing process
- Alternative splicing
- Controls which extrons are getting spliced together
Spliceosome composition
1) Spliceosomal proteins and small nuclear RNAs (snRNAs)
2) snRNAs are called U RNAs (uracil rich)
- U1, U2, U3..
3) snRNAs recognize the intron elements
- RNA-RNA base pairing
Spliceosome assembly and function
Splice site and branch point recognition
SNPs U1 and U2 bind to the 5` splice site and internal A
The U4-U5-U6 complex joins
Intron folding
Makes catalytically active
Splicing step 1
First splicing reaction: one intron end attaches to A
Splicing step 2
Second splicing reaction: other intron end cleaved' exons join
Summary
Spliceosome recognizes the conserved intron elements
Splice site definition
Different snRNPs assemble at different stages
Dynamic complex
Only a subset of snRNPs are present during the trans-esterification reactions
(U2, U5 and U6)
mRNA in eukaryotes:
-Sequences present in the primary transcript may not be present in the mature mRNA.
-Sequences present on the 5' and 3' ends of the mature mRNA were not present in the DNA
template.
-Removal of introns generates a lariat intron product.
-Exons from a single gene may be spliced (joined) together in different combinations.
Cotranscriptional processing of RNA in eukaryotes
The phosphorylation form of RNA polymerase II interacts with the complexes involved in capping, splicing,
and polyadenylation
-efficiency of mRNA production
a) Capping
- Interacts with phosphorylation
- Activates catalytic activity
b) Splicing
- Phosphorylation status changes
c) cleavage and polyadenylaiton
- phosphorylation status changes again (functionality)
Self-splicing introns
-spliceosomal introns require the spliceosome for catalysis
-other intron classes can splice themselves
no protein factors required
-first identified in a rRNA gene in Tetrahymena (Cech, 1989)
also found in mitochondrial, chloroplast and bacterial genes
Self-splicing reaction (autocatalytic)
Group I intron
Free GTP is used as the first nucleophile
-attacks the phosphodiester bond at the 5` splice site using the 3` hydroxyl
-now have a new 5` end which has an added G from GTP
-5` exon joined to 3` end
Products are ligated exons and a linear intron
The RNA World
Walter Gilbert - coinventor of DNA sequencing; looked at introns
- RNA can store genetic information (viruses, mRNA) and RNA can undergo enzymaticlike
reactions
- Catalytic RNAs are called ribozymes (RNA-based enzymes)
Therefore, RNA can act both like DNA genetic material and as protein
(storing of information and enzymatic activity)
However, both DNA and proteins separately are better at their roles
DNA is more stable than RNA
Proteins can have more enzymatic variety than RNA
RNA could have been the ancestral genetic material
Small RNAs were the breakthrough of the year (non-coding RNAs)
Small RNAs control gene expression at all levels:
DNA - structure of chromation
RNA - both informational an functional RNAs affected
Protein synthesis
Similar suppression phenotypes can be induced by injecting cells with double-stranded RNA
molecules
Eg) pigment inhibition in petunia flowers
Different strands serves as templates for different genes but only one strand of RNA is produced. Any gene
in the chromosome is only producing one RNA strand.
Trans-genetics
-unusual insertion sites for introduced genes may generate double-stranded RNA
mRNA produced from the transgene, and gene adjacent to the transgene also has a promoter so another
RNA strand is produced
RNA Interference (RNAi)
Gene silencing mediated through small RNA molecules
Double-stranded RNA acts as the trigger mechanism
Precursor - double stranded RNA
RNA cleavage occurs
RNA interference
A way to silence genes using small RNA molecules
We looked at the mechanism where you can silence a gene at the molecular level
Example: petunia flowers, suppression of purple pigment artificially
Pathway starts with double stranded RNA molecules
RNAi - interfering RNA
Trans-genetics
Unusual insertion sites for introduced genes may generate double-stranded RNA
mRNA
---Transgene------Gene---
--Promoter--
siRNAs - small interfering RNAs
Mechanism of action of RNAi
Dicer - an endoribonuclease
A double-strand specific
The enzyme "dicer" cuts the double stranded RNA into smaller sections
One of the siRNAs will be incorporated into risc complex (RISC)
RISC - RNA induced silencing complex:
-RISC binds to the single-stranded siRNA and then binds to the mRNA
-mRNA cleavage and degraded
Therefore inactivates mRNA
Gene is silenced
RISC - RNA induced silencing complex
Binds a single-stranded siRNA and binds the mRNA
mRNA cleavage
need a strand of DNA that is complementatary to the mRNA
In order for this to happen, one strand of the double-stranded siRNA must be complementary to the target
mRNA
Specific base-pairing between an siRNA and mRNA target
mRNA cleaved and inactivated
The RNAi technique is particular useful because:
1) Gene replacement or knockout is not required to eliminate gene expression
- RNAi is fast to employ in a lab setting and can be used to inactivate gene expression at different
development stages
2) Specific - target only one mRNA for destruction
- possible treatment for viral infections or cancer cells?
Additional Chapter 8 Problems: # 6 , 15
Chapter 9 Proteins and their Synthesis
-Protein synthesis occurs through the process called translation
-Translation occurs on RNP particles called ribosomes
Year 2000 - first high resolution 3D structures solved for ribosomes:
How does the ribosome work?
How is the ribosome inhibited - antibiotics?
The binding of a drug to the ribosome prevents translations
Eg) -Large ribosomal subunit
Erythromycin (drug) binds to the middle of the ribosome
The binding site is the exit point of the amino acid chain - channel where the polypeptide leaves the
ribosome
Translation components
-Ribosome - RNA complex
- catalyze peptide bond formation
-Transfer RNAs (tRNAs)
- bring amino acids to the ribosome
- decode information on the mRNA
-Messenger RNA
.
Protein Structure
-amino acids with different properties
-protein sequence and 3D shape influences function
-molecules that can be bound
-cellular location
-protein-protein interactions
-Protein 3D structure is stabilized by:
-ionic interactions
-H bonding
-hydrophobic and Van der Waals forces
Amino Acid Structure
There are 20 different standard amino acids in proteins
Side chain R group is different for each one
Contains a central carbon, an amino group, and a carboxylic acid group.
Different side chain (R group) properties:
-charged

-Protein = chain of amino acids linked by peptide bonds = polypeptide chain
Peptide bond: carbox joined to amino
Synthesized from the amino end towards the end that will be the carboxyl end
Atoms in the peptide bond are co-planer
Oxygen and nitrogen atoms of the peptide bond can engage in H-bonds with other amino acid positions
The peptide bond is stable at physiological pH
Levels of protein structure
1. Primary structure = linear sequence (order) of amino acids within the polypeptide
2. Secondary structure - arises from interactions between "nearby" amino acids (e.g. H-bonds form an
-helix). The amino acids aren't necessarily directly adjacent but are relatively close in the
linear sequence.
3. Tertiary structure - the overall 3-d shape of the entire protein. Does the protein adopt a globular
structure or is it elongated? Where are the precise positions of all the atoms of the protein located
in 3-d space? How are all the secondary structural elements positioned relative to each other?The
illustrated hemoglobin complex contains: 2 subunits and 2 subunits (4 polypeptides total).
Therefore, this structure is a heterotetramer.
4. Quaternary structure - interactions between multiple polypeptides E.g. dimer = two polypeptides,
tetramer - 4 polypeptides
Homo - identical (protein products from same gene) Hetero - different (from at least 2 genes - different
polypeptides)
The Genetic Code
How is the sequence in DNA converted into amino acid sequence in proteins?
- How is mRNA read as a template during translation? 1)
How many nucleotides specify one amino
acid in a protein? 2) What is the directionality of reading an mRNA?
The Genetic Code The length of mRNA sequence that specifies one
amino acid = Codon Codon length
1 nt = 4 possible combinations (4 amino acids) 2 nt = 4 X 4 = 16 amino acids could be specified 3 nt = 4 X 4
X 4 = 64 possible amino acids could be
specified
Therefore, codons of 3 nt length contain enough information to specify the 20 different standard amino
acids found in protein structure
Overlapping versus non-overlapping genetic codes
Overlapping - each nucleotide is part of more than one codon Non-overlapping - each nucleotide is only
part of one codon
Mutants and Suppressors demonstrate a non-overlapping triplet code (Crick and Brenner, 1961)
Observation #1: A single substitution mutation (1 nt change) only ever changes 1 amino acid within a
protein.
Therefore, code is non-overlapping
Observation #2: Only certain combinations of addition and deletion mutants suppress mutant phenotypes
(re-establish protein functionality) Additions in multiples of 3
Deletions in multiples of 3 or equal numbers of additions and deletions
For example portion of mRNA sequence
wild - type
5' --------------- GAU CGA CUU GGA AAA UUU GCA CCA ------------------3'
1 addition
Asp- Arg- Leu- Gly - Lys - Phe - Ala - Pro
mutant
5' --------------- GAU UCG ACU UGG AAA AUU UGC ACC A -----------------3'
Asp- Ser- Thr- Trp - Lys - Ile - Cys - Thr
Changed portion of mRNA sequence
SYNTHETIC POLYRIBONUCLEOTIDES DIRECT SYNTHESIS OF POLYPEPTIDES IN A CELL-FREE
SYSTEM
(Nirenberg & Matthaei, 1961)
mRNA template - UUUUUUUUUUUUUUUIn vitro peptide synthesized polyphenylalanine: (Phe)n
polylysine: (Lys)n Therefore, UUU is the codon for phenylalanine (Phe)
AAA is the codon for lysine (Lys)
Likewise:
poly(G):
Gly (only)
poly(Gly)
GGG = Gly poly(C):
Pro (only)
poly(Pro)
CCC = Pro
-AAAAAAAAAAAAAAAAACHEMICALLY-SYNTHESIZED POLYRIBONUCLEOTIDES OF DEFINED SEQUENCE DIRECT THE
SYNTHESIS
OF POLYPEPTIDES OF DEFINED SEQUENCE (Khorana)
e.g., poly(UG)
... UGU|GUG|UGU|GUG|UGU|GUG|UGU|GUG ... ... Cys-Val-Cys-Val-Cys-Val-Cys-Val
...
The same alternating sequence of two amino acids
(Cys and Val) results regardless of where translation begins.
USE OF SYNTHETIC POLYRIBONUCLEOTIDES WITH REPEATING SEQUENCES TO DECIPHER
THE CODE
This example shows how polypeptides derived from the (AAG)n polymer were used to confirm the triplet
code and help to identify the codons. The (AAG)n polymer can specify three different polypeptides,
depending on which reading frame is used.
Chapter 9 Practice Problems: 1, 2, 3, 5, 7, 9, 16, 17, 27, 28
The Genetic Code (written as RNA)
Standard Genetic Code - the vast majority of organisms use the same codes for amino acids
Chain termination (`stop`) codons are shown in orange
UAA
UAG
UGA
The usual intial (`start`) codon is shown in green
AUG
For Example
5`
CAC AGU
3`
His-Ser
NH3+ Hist - Ser - COO-
Degeneracy and the Genetic Code
The genetic code is:
Unambiguous: one codon, one amino acid
Eg) methionine encoded by AUG
Degenerate: one amino acid, more than one codon
Transfer RNA (tRNA) is the adaptor molecule
Base-pairing
mRNA codon: tRNA anti-codon
At least 20 different tRNAs
Amino acids are temporarily covalently attached to tRNAs - "charging"
The accuracy of attaching the correct amino acid (cognate) to the correct tRNA is critical for ensuring
translation accuracy
Accuracy - same amino acid to same tRNA species - called the cognate species
The structure of tRNA
-many base pairing interactions
-small, but highly structured
-amino acids are always added onto the very 3` end of the tRNA
-the other end interacts with the mRNA codon sequence
called the tRNA "anticodon"
Specificity
-Binds only one amino acid (cognate)
-Binds only cognate tRNA(s)
-ATP is required during the charging reaction
Aminoacyl-tRNA synthetase is specific for an amino acid
Eg) alanine
Binding site for alanine
Binding site for tRNAAla
Enzyme joins carboxylic acid carbon of alanine onto the 3` end of tRNA
Requires ATP
The structure of tRNA
Four stems of base-pairing interactions in a 2-D cloverleaf representation
Inverted L shape in the 3-D representation
-3` end is amino acid attachment site
-DHU loop
-TyC loop
-Anticodon loop
Two superimposed yeast tRNAs
-notice how different species of tRNAs have the same path of phosphodiester backbone; this nearsuperimposable structure must be highly stable
-tRNA structure is different in anticodon region, and A.A. binding site
Genetic code is degenerate
Amino acid serine is specified by 6 different codons.
How does the cellular tRNA set relate to the degeneracy?
Eg) Different tRNAs that can service codons for Serine
tRNA
Anticodon
Codon
tRNASer1
AGG+ wobble
UCC
..
How does the cellular tRNA set relate to the degeneracy?
1) More than one tRNA species can be charged with the same amino acid
Isoacceptors
Most organisms contain at least 3 different tRNAs that are charged with serine: tRNASer1
2) A single tRNA species can recognize more
Wobble allows one tRNA to recognize multiple codons
RNA can engage in non-canonical base pairs such as G-Upairs
This tRNA can recognize either 5`UCC3` or 5`UCU3` codons
Wobble position
Non-colonical? pairings; G can base pair to a U
This is restricted to only the anticodon's 5` end, so the mRNAs 3` codon end
This explains the genetic table; Wobble pairing explains patterns why multiple codons can encode
for the same amino acid
Wobble poisiton is 5` in the anticodon
3` in the codon on the mRNA (third position)
Codon-Anticodon Pairing Allowed by Wobble Rules:
5` end of anticodon tRNA
3` end of codon mRNA
G
C or U
C
G only
A
U only
U
A or G
I
U, C, or A
I= Inosine, a modified base
Modified nucleotide forms from converting an A into an I, present in tRNA
Similar structure to guanine
Ribosome structure
-small subunit (SSU) and large unite (LSU)
-conservation of structure - particularly in the ribosomal RNAs (rRNA)
Prokaryotic vs. Eukaryotic Ribosomes
-prokaryotic are smaller
-fewer ribosomal proteins and shorter rRNAs
Prokaryotic Ribosomes
-the 70 S ribosome
-two subunits, small (30 S) and large (50 S)
-small 16S rRNA and 21 proteins only found in the small subunit
-large 23S rRNA plus 5S rRNA plus 31 proteins
Eukaryotic Ribosome
-the 80S ribosome is larger
-small ribosomal subunit is 40S
-large subunit is 60S
-produces more proteins; has conterparts to the prokaryotic rRNAs
-Large produces three rRNAs
Ribosomal RNAs (rRNAs) are highly structured
-engage in lots of base pairing
-intramolecular base-pairing interactions
-secondary structure
-RNA helices
-conserved tertiary structure
-3D structural conservation (prokaryotes and eukaryotes)
-binding sites for ribosomal proteins, mRNA,
..
Secondary structure of prokaryotic 16S (small subunit) rRNA
-rRNA folds up by intramolecular base pairing
Short range interaction
-stem-loop element
Long range interaction
-base-pairing between distant sites
Ribosome subunit roles
-Small subunit
-bind the mRNAs
-decoding site
-Large subunit
-catalyze peptide bond formation peptidyl transferase activity
Key sites of interaction in the ribosome
Computer model
50S large subunit
-polypeptide exit tunnel
-peptidyl transferase center
30S small subunit
-mRNA binding site
3 tRNA binding sites:
E site, P site, and the A site
As the polypeptide grows, (from amino terminus to carboxy terminus) it becomes too large for the
ribosome and must be released
Key sites of interaction in the ribosome
tRNA movement
A site P site E site
Opposite direction of motion of the ribosome along the mRNA strand
Schematic model
When we form a peptide bond, always have two tRNAs binded
tRNA binding sites
A site - Amino acyl site
Next incoming charged tRNA binds here
-next amino acid to be added
Incoming tRNA that matches the mRNA enters ribosome at A site
P site Peptidyl site
tRNA temporarily carrying the elongated peptide chain resides here
E site - Exite site
tRNA temporarily resides here just prior to leaving the ribosome
Deacylated tRNA released from the E site
Initiation of translation
1. Binding of mRNA first to the small subunit (SSU)
2. Recognition of a start codon
a. Usually AUG
b. Utilizes a specialized initiator tRNA
i. Prokaryotes - tRNAfMet
ii. Eukaryotes - tRNAiMet
3. Requires protein translation factors called initiation factors
How is the correct AUG codon chosen?
In prokaryotes, sequences upstream of an AUG start codon define the start site by binding to the
small subunit of the ribosome.
We use the plank sequence (upstream) which makes interactions to serve as a binding site.
Shine Dalgarno Sequence
Upstream of an AUG codon
Preference with a purine nucleotide
Small subunit of mRNA positioned upstream of AUG codon
Base-pairing between Shine Dalgarno sequence in mRNA and anti-Shine-Dalgarno sequence at
3`
end of small subunit RNA
Translation initiation in prokaryotes
30S ribosomal subunit
+ initiation factors
1. Keep subunits separated
IF2-fMet-tRNAf + mRNA
Gives
fMet and IF3
2. Recruit the initiatior tRNA and position it at a start codon
3. Binding of the large subunit
a. Initiation factor release
b. tRNAfMet is positioned in the P site
Once the initiation process occurs, it is essential that the initiator tRNA is in P site
Translation initiation in Eurakyotes
Difference: no Shine-Daglarno sequence
Instead, take advantage of 5` cap structure for recognition during initiation
1. Recognition of the mRNA cap structure
2. Scanning to find an appropriate start codon
a. AUG and flanking nucleotides
Not necessarily going to be close to the 5` end
Ribosome moves 5` to 3`
b. Energy dependent
Requires hydrolysis of ATP
3. Binding of the large subunit
a. Initiation factors released
b. tRNAfMet is positioned
Translation Elongation
-Covers all steps where we make peptide bonds
-Progressive creation of peptide bonds until a stop codon
Cycle:
1. aminoacylated-tRNA binds to the A site of the ribosome
2. formation of peptide bonds through peptidyl transferase reaction
3. Translocation
a. Ribosome re-positions itself
b. .
Steps in translation elongation in prokaryotes
-tRNA is binded to the P site
-Translocation factor EF-Tu brings aminoacylated tRNA to the ribosome
-Peptidyltransferase reaction moves peptide (or first amino acid) to the A site tRNA
-Elongation factor EF-G displaces A site tRNA
GTP dependent
mRNA moves through the ribosome
Translocation Termination
1. Stop codon not recognized by a tRNA anticodon
a. Recognized by a release factor
Prokaryotes: RF1, RF2 (specifically to recognize stop), and RF3
2. polypeptide is released from tRNA and ribosome
3. ribosome subunits dissociate from each other and mRNA template
Termination of translocation
-Release factor recognizes a stop codon
assisted by other RF's and protein factors
hydrolysis breaks the bond joining the polypeptide to tRNA
Post-translational processing events
-Covalent additions of chemical groups or peptides
phosphates, sugars, ubiquitin
covalent bonds are formed
-Protein sorting and targeting
signal sequences
-Protein folding (also co-translational)
chaperonin complexes associate with unfolded proteins to allow folding
in some cases the chaperonin complex can occur at the ribosome
and therefore be either co-translational or post-translational
Phosphorylation
-protein kinases use ATP as a substrate and add it onto a specific amino acid position to convert it from
inactive to active enzyme
Phosphorylation Regulates
- enzymatic activity
- protein-protein interactions
or the association of protein complexes
- protein-nucleic acid interactions
phosphates are charged groups that can interact with acids
- cellular localization
Ubiquitin may target a protein for degeneration
Ubiquitin is a highly-conserved small protein added onto lysine side chains
Ubiquitin chain is degraded Ubiquitin is attached to a different protein Ubiquitin protein is
formed addition of multiple ubiquitins (ubiquitin chain) is a signal for protein digestion called a
26S protease protease degrades proteins results in oligopeptides and ubiquitin chain
Signal sequences (protein sorting)
- short peptide sequences within the protein with distinctive chemical properties
- used to direct proteins to the nucleus, mitochondria, chloroplasts, plasma membrane, for excretion
- signal sequence may be removed (cleaved) or retained
Signal sequences target proteins for secretion
Signal recognition particle - an RNP - directs the ribosome to the translocation complex in the endoplasmic
reticulum membrane