IB BIOLOGY
TOPICS 4 AND 10: GENETICS
I. DNA Structure
A. Elements are carbon, hydrogen, oxygen, nitrogen and phosphorus
B. Monomers are called nucleotides
Figure 1. Structure of a nucleotide and the nitrogenous bases.
http://www.yellowtang.org/images/structure_of_nucleo_c_la_784.jpg
1. Nucleotide is made of a sugar (ribose or deoxyribose), one or more
phosphate groups and a nitrogenous base
*note that a nucleoside is the sugar and the nitrogenous base, no
phosphate groups
a. The carbon with the nitrogenous base is called the 1’ (one prime) carbon
b. The carbon with the H in deoxyribose or the OH in ribose is called the 2’
(two prime) carbon
c. The carbon with the phosphate group is called the 5’ (five prime) carbon
d. When 2 nucleotides join, the 5’ phosphate of one nucleotide bonds wtih
the 3’ carbon of the next nucleotide
e. Think of the 5’ carbon as being the front (“f” for five and “f” for front) of
a train with other nucleotides joining to the 3’ carbon at the back of the
train (usually more cars are attached to the back end of the previous
car, not the front end)
*helpful for the processes of replication and transcription
f. Nitrogenous bases are either purines or pyrimidines
i. purines
a. adenine and guanine
b. structure has 2 rings
ii. pyrimidines
a. cytosine, thymine and uracil
b. structure has 1 ring
C. Arrangement of nucleotides
Figure 2. Complementary base pairing with detail of hydrogen bonding.
http://www.starsandseas.com/SAS%20OrgChem/SASNucleic.htm
Figure 3. Space-filling model of DNA.
0560|00590|00010|00020|00030|00040|00050|00060|00070|00120|00080|00090|00100|00110|01000|02000|03000|04
Figure 4. A nucleotide represented with simple shapes as outlined in the IB syllabus.
Figure 5. Part of one strand of DNA represented with simple shapes as outlined in IB
syllabus.
1. 2 chains/strands of nucleotides
a. Each chain/strand has alternating deoxyriboses and phosphates that act
as a “backbone” with the nitrogenous bases sticking out between the
chains
b. Chains held together by hydrogen bonds between the nitrogenous bases
i. complementary base pairing: a purine always bonds to a pyrimidine
a. A always binds to T
b. C always binds to G
c. important to replication, transcription and translation
c. The 2 strands run anti-parallel to each other: the 3’ end of one strand is
at the same end as the 5’ end of the other strand
II. Chromosome Structure
Figure 6. Levels of structure of chromatin.
http://www.nature.com/scitable/resource?action=showFullImageForTopic&imgSrc=18847/pierce_11_5_FULL.jpg
Figure 7. Levels of structure of chromatin and reduction in length.
http://www.biochem.arizona.edu/classes/bioc462/462a/NOTES/Nucleic_Acids/nucacid_structure.html
Figure 8. Electron micrograph of chromatin from chicken red blood cell.
http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/N/Nucleus.html#Nucleosomes
Question: What is odd about the preceding Figure? Read the title carefully!
___________________________________________________________________
A. Made of units called nucleosomes
1. DNA wrapped around 8 histone proteins and held together by another
histone protein
a. Each nucleosome has about 165 DNA base pairs
b. Nucleosomes join together to form chromatin
c. Each chromosome contains 100’s of 1000’s of nucleosomes, joined by
the DNA that runs between them (about 20 base pairs)
i. called linker DNA
2. Nucleosome formation helps to make the DNA compact enough to fit in the
nucleus
a. allows supercoiling of the DNA
i. Regulates gene transcription because the RNA polymerase cannot
easily access the gene
a. the DNA must be partially uncoiled at the relevant gene for
transcription to occur
i. done by modifying the histones: adding acetyl, methyl or
phosphate groups; displacing histone with chromatin
remodelling complexes which exposes the underlying DNA
sequences to polymerases etc.
Figure 9. Supercoiling of DNA.
http://oregonstate.edu/instruction/bb492/lectures/StructureIII.html
For some information about the size of each chromosome and what genes are on it:
http://genomics.energy.gov/gallery/chromosomes/gallery-01.html
III. Meiosis
Figure 10. The stages of meiosis .
http://www.britannica.com/bps/media-view?1
A. Differences between meiosis and mitosis:
1. The number of chromosomes in the daughter cells is ½ of the number of
chromosomes in the parent cell
2. 4 daughter cells are produced
3. (i.e. a cell that is 2n (diploid) produces 4 cells that are n (haploid))
4. There are 2 divisions (meiosis I and meiosis II)
5. In metaphase I, the chromosomes line up at the equator in pairs but in
metaphase II, the chromosomes line up at the equator in single file
B. Formation of Sex Cells/Gametes
Figure 11. Spermatogenesis and oogenesis.
http://www.bio.miami.edu/dana/104/gametogenesis.jpg
C. Crossing Over
Really fast animation: http://www.tokyo-med.ac.jp/genet/anm/mimov.gif
Figure 12. Crossing over.
1. Occurs during prophase I
2. As the chromatids are tightly entwined, produces stress on the DNA of the
chromatids and they can break
3. Repair is not perfect because the genetic sequences of homologous
chromosomes are so similar so that a fragment of one chromatid is
rejoined with the broken chromatid of the neighbouring chromosome
rather than with the original chromatid
4. The site of the crossing over is called a chiasma (plural chiasmata)
D. Non-disjunction
1. Non-disjunction: an error in chromosome sorting during either meiosis I or
meiosis II
a. In meiosis I, result is 2 daughter cells that have 2 copies of the
chromosome and 2 cells that are missing the chromosome altogether
Figure13. Non-disjunction in meiosis I and the result of fertilization.
http://www.medgen.ubc.ca/robinsonlab/mosaic/intro/tri_how.htm
b. In meiosis II, the sister chromatids do not separate and travel to the
same cell
Figure 14. Non-disjunction in meiosis II and the result of fertilization.
http://www.medgen.ubc.ca/robinsonlab/mosaic/intro/tri_how.htm
2. Autosomal trisomies that can produce a viable offspring in humans (other
trisomies do not produce viable offspring)
a. Down syndrome/Trisomy 21
i. affects 1:700 children
ii. produces characteristic facial features, short stature, heart defects,
susceptibility to respiratory disease, shorter lifespan, prone to
Alzheimer’s and leukemia, possible sexual underdevelopment and
sterility, some degree of intellectual challenge
iii. correlated with age of mother but can also be the result of nondisjunction in father
b. Patau syndrome/Trisomy 13
i. 1:5 000 live births
ii. serious eye, brain and circulatory defects, cleft palate
iii. children rarely live more than a few months
c. Edward’s syndrome/Trisomy 18
i. 1: 10 000 live births
ii. almost every organ system affected
iii. children rarely live longer than a few months
3. Non-disjunction of sex chromosomes
a. Klinefelter syndrome/XXY males
i. have male sex organs
ii. unusually small testes, sterile
iii. some feminine body characteristics such as breast enlargement
iv. normal intelligence
b. Supermale syndrome/XYY males
i. somewhat taller than average
ii. often have below normal intelligence
iii. once thought to be more likely to be criminally aggressive but that
has been disproven
c. XXX females/trisomy X
i. 1: 1 000 live births
ii. healthy and fertile
iii. usually can not be distinguished except by karyotype
d. Monosomy X/Turner’s syndrome/XO females
i. 1:5 000 live births
ii. the only viable monosomy in humans
iii. do not mature sexually and are sterile
iv. short stature
v. normal intelligence
IV. Theoretical Genetics
A. Essential Definitions
1. Gene: a heritable factor that controls a specific characteristic. Each type of
chromosome has a number of genes located on it. Since human
somatic cells are diploid, each human cell has two copies of each
type of chromosome and therefore two copies of each gene. These
copies may be of the same form or different forms from each other.
2. Allele: one specific form of a gene differing from other alleles by only one
or a few bases and occupying the same gene locus as other alleles
of the gene
3. Genome: the whole of the genetic information of an organism
4. Genotype: the alleles of an organism
5. Phenotype: the characteristics of an organism, controlled by genotype
6. Dominant allele: an allele that has the same effect on the phenotype
whether it is present in the homozygous state or
heterozygous state
7. Recessive allele: an allele that only has an effect on the phenotype when
present in the homozygous state
Table 1. The dominant and recessive traits that were studied by Mendel.
Pea Plant Characteristic
Seed shape
Seed colour
Flower colour
Pod shape
Pod colour
Flower position
Plant height
Dominant Trait
Round
Yellow
Purple
Inflated
Green
Axial
Tall
Recessive Trait
Wrinkled
Green
White
Constricted
yellow
Terminal
Short
8. Codominance/Codominant alleles: pairs of alleles that both affect the
phenotype when present in a
heterozygote
9. Incomplete dominance*: term used to refer to the affect of two
codominant alleles when their effects are blended
in the phenotype e.g. pink flowers in snapdragons
10. Partial dominance*: term used to refer to the affect of two codominant
alleles when both of their effects are separately
expressed in the phenotype e.g. roan fur in cattle
*According to IB, these terms are no longer used.
11. Locus: the particular position on homologous chromosomes of a gene
12. homozygous: having two identical alleles of a gene
13. heterozygous: having two different alleles of a gene
14. carrier: an individual that has one copy of a recessive allele that causes a
genetic disease in individuals that are homozygous for the allele
15. test cross: testing a suspected heterozygote by crossing it with a known
homozygous recessive
16. Punnett Square/Grid: a device used in genetic analysis, does not have to
be a square but can be rectangular
Gametes of Parent 1
Gametes of Parent 2
Allele in first
type of gamete
P2A1
Allele in second
type of gamete
P2A2
Allele in first
type of gamete
P1A1
Genotype of
offspring 1
(P1A1, P2A1)
Genotype of
offspring 3
(P1A1, P2A2)
Allele in second
type of gamete
P1A2
Genotype of
offspring 2
(P1A2, P2A1)
Genotype of
offspring 4
(P1A2, P2A2)
17. Sex chromosome: chromosomes that determine gender in humans,
denoted X and Y. Females have 2 X chromosomes,
males have 1 X chromosome and one Y chromosome.
18. Sex-linked trait: a trait governed by a gene that is found on a sex
chromosome, usually the X chromosome.
For HL:
19. Autosome: any chromosome that is not a sex chromosome
20. dihybrid cross: a cross that involves 2 traits (genes)
21. (Mendel’s) Law of Independent Assortment: the inheritance of an allele of
one gene does not influence which allele of a second gene is inherited
22. linkage: the situation where two or more genes are found on the same
chromosome and are “linked”; therefore certain combinations of
alleles tend to be inherited together when predicting the result of
mating
For a longer, more detailed explanation:
http://www.nature.com/scitable/topicpage/some-genes-are-transmitted-tooffspring-in-6524945
23. linkage group: a group of genes that is found on the same chromosome;
therefore the gene combinations of each parent tend to be
inherited together (i.e. do not follow the Law of
Independent Assortment)
Figure 13. Some linked genes in Drosophila (fruit fly). (Note: I have not checked
what the original source of this figure is.)
http://www.biologycorner.com/APbiology/inheritance/12-2_gene_linkage.html
Here is a worked example: http://www.biologycorner.com/APbiology/inheritance/122_gene_linkage.html
Example: A fly that is heterozygous for long wings (Ll) and heterozygous for long
aristae (Aa) is crossed with another fly of the same type. AaLl x AaLl. In both cases
the dominant allele is located on the same chromosome.
24. recombinants: can refer to chromosomes that result from crossing over,
offspring that have genotypes that are different from
either of the parents’ genotypes or to organisms that have
been genetically engineered to have DNA from 2 different
species
25. crossing over: the phenomenon where homologous chromosomes
exchange fragments during either meiosis I or meiosis II.
For animation of definitions 20, 21, 24, and 25:
http://bcs.whfreeman.com/thelifewire/content/chp10/1002s.swf
Figure 14: An example of the results of crossing over in fruit flies.
http://www.biologycorner.com/APbiology/inheritance/12-2_gene_linkage.html
Note that the proportion of offspring that result from the recombinant chromosomes
is small.
26. polygenic inheritance: describes the inheritance of a trait that is controlled
by more than one pair of genes and usually results
in continuous variation in the trait. Examples
include human skin colour, human height and
wheat kernel colour, plant height in tobacco, Rh
factor in humans
Table 2. Polygenic inheritance in people showing a cross between two mulatto
parents (AaBbCc x AaBbCc). The offspring contain seven different shades of skin
color based on the number of capital letters in each genotype.
http://waynesword.palomar.edu/lmexer5.htm
*This website has other examples of polygenic traits with tables showing the
combinations of alleles as well.
Multiple Gene (Polygenic) Inheritance
Gametes ABC
ABc
AbC
Abc
aBC
aBc
abC
abc
ABC
6
5
5
4
5
4
4
3
ABc
5
4
4
3
4
3
3
2
AbC
5
4
4
3
4
3
3
2
Abc
4
3
3
2
3
2
2
1
aBC
5
4
4
3
4
3
3
2
aBc
4
3
3
2
3
2
2
1
abC
4
3
3
2
3
2
2
1
abc
3
2
2
1
2
1
1
0
B. Performing Genetic Crosses
*To learn how to perform crosses, please follow the examples below.
Example 1: Complete dominance.
In humans, the allele for the ability to roll one’s tongue in the shape of a “U”
is dominant over the allele for an inability to roll one’s tongue. What is the
expected proportion of genotypes in the offspring if two parents are both
heterozygous for this gene?
Step 1: Identify or choose appropriate symbols to represent the alleles. It is usually
appropriate to choose the first letter of the dominant trait. The dominant
trait is represented by the capital letter and the recessive trait is
represented by the lowercase letter.
Let R represent the allele for ability to roll one’s tongue and r represent the allele for
the inability to roll one’s tongue.
Step 2: Determine the genotypes of the parents.
Since both are heterozygous, their genotypes must be Rr and Rr.
Step 3: Determine the alleles that will be found in each parents’ gametes and write
them into a Punnett Square. (Note that Punnett Squares may become
unnecessary as you become more comfortable with genetic analysis.)
Recall the formation of gametes involves the process of meiosis. Therefore,
each gamete will have only one copy of each type of chromosome and only
one allele for each trait.
Therefore the male parent will produce two kinds of gametes: one with R and one
with r. The same is true of the female parent.
Gametes of Parent 1
Gametes of Parent 2
Allele in first
type of gamete
R
Allele in second
type of gamete
r
Allele in first
type of gamete
R
Genotype of
offspring 1
RR
Genotype of
offspring 3
Rr
Allele in second
type of gamete
r
Genotype of
offspring 2
Rr
Genotype of
offspring 4
rr
Step 4: Address the question. Frequencies/proportions can be expressed as
fractions, decimals, percents, or ratios. Be careful that you read the
question carefully, your answer matches the question and the method you
choose is acceptable based on the wording of the question.
¼, 0.25, 25% 1:4 offspring will have the genotype RR.
½ , 0.5, 50% 1:2 offspring will have the genotype RR.
¼, 0.25, 25% 1:4 offspring will have the genotype RR.
Example 2: Codominance with incomplete dominance.
In snapdragons, the alleles for red and white flower colour are codominant.
The heterozygous genotype produces a phenotype of pink flowers. A plant
that produces white flowers is crossed with a plant that produces pink
flowers. What percentage of the offspring would produce pink flowers?
Step 1: In cases of codominance, it is common to use capital letters to represent
each allele.
Let R represent the allele for red flowers and W represent the allele for white
flowers.
Step 2:
The genotype of the plant that produces white flowers is WW and the genotype of
the plant that produces pink flowers is RW.
Step 3: Notice that since the plant that produces white flowers is homozygous, it is
actually unnecessary to complete the second column (Gametes of Parent 1,
Allele in second type of gamete).
One parent only makes gametes with W, the other parent makes two types of
gametes, one with R and one with W.
Gametes of Parent 1
Gametes of Parent 2
Allele in first
type of gamete
R
Allele in second
type of gamete
W
Allele in first
type of gamete
W
Genotype of
offspring 1
RW
Genotype of
offspring 3
WW
Allele in second
type of gamete
W
Genotype of
offspring 2
RW
Genotype of
offspring 4
WW
Step 4:
Therefore 50% of the offspring will produce pink flowers.
Example 3: Codominance with partial dominance.
In cattle, there are two alleles for fur colour: one allele for white and another for red
(reddish brown really). The heterozygous genotype results in a cow that makes both
types of fur, red and white, which is called roan. What fraction of the offspring from
two roan animals would be expected to have roan fur?
Step 1:
Let R represent red fur and W represent white fur.
Step 2:
Since both parents have roan colouring, they are both heterozygous or RW.
Step 3:
Both parents make gametes of two types: R or W.
Gametes of Parent 1
Allele in first
Allele in second
Gametes of Parent 2
Allele in first
type of gamete
R
Allele in second
type of gamete
W
type of gamete
R
Genotype of
offspring 1
RR
Genotype of
offspring 3
RW
type of gamete
W
Genotype of
offspring 2
RW
Genotype of
offspring 4
WW
Step 4:
2 of the 4 offspring have the RW genotype and will have roan fur. As a fraction, this
is ½.
Example 4: Codominance with partial dominance and multiple alleles.
* Note that beginning with this example, the descriptions of the sections of the
Punnett Square will be omitted.
In humans, there are 3 alleles for blood type: IA, IB, and i. IA and IB show partial
dominance and both are dominant over i. As a result, there are 4 blood types** as
in the table below:
Table 3: Human blood types (phenotypes) and genotypes.
Phenotype
Blood Type A
Blood Type B
Blood Type AB
Blood Type O
Possible Genotypes
IA IA, IAii
IB IB, IBi
IA IB
ii
** IA codes for a protein found on the cell membrane of red blood cells called
antigen A, IB codes for a protein found on the cell membrane of red blood cells
called antigen B while i does not code for either of these proteins.
Would it be possible for a couple to have a child with Blood Type O if one parent has
Blood Type A and one parent has Blood Type O?
Steps 1 and 2: In this situation, use the standard notations for the alleles.
The parent with Blood Type A: IAIA or IAi
The parent with Blood Type O: ii
Step 3:
In this case there are 2 possible situations. You should look at both in detail in order
to answer the question. Note that once again, if a parent is homozygous, some of
the columns/rows in the Punnett Square are not necessary.
If parent with Blood Type A is IAIA:
i
i
IA
IA i
IA i
IA
IAi
IAi
IA
IA i
IA
i
ii
ii
If parent with Blood Type A is IAi:
i
i
Step 4:
It is possible for a parent with Blood Type A and another parent with Blood Type O
to have a child with Blood Type O but only if the parent with Blood Type A is
heterozygous.
Example 5: Hemophilia, a sex-linked trait.
Hemophilia is a bleeding disorder that results in a reduced ability/speed of blood
clotting. There is more than one form of haemophilia but the two most common
forms are both sex-linked traits with the gene found on the X chromosome. The
allele that codes for the defective protein is recessive.
What is the probability that a woman who is a carrier of the haemophilia allele and a
normal man will have a child that is a haemophiliac? What is the probability that a
son from this relationship will be a haemophiliac?
Step 1:
Let XH represent the allele for normal blood clotting and Xh represent the allele for
haemophilia (reduced ability of blood to clot).
Step 2:
Woman: XH Xh
Man: XH Y
Step 3:
Woman’s gametes: XH and Xh
Man’s gametes: Xh and Y
XH
XH
XH
XH
XH
Y
Xh
Xh
XhY
XH
Step 4:
Only the child with the genotype XhY will have haemophilia so the probability of
having a child with haemophilia is ¼ or 25% or 0.25.
Of the possible sons, the probability of one being a haemophiliac is ½ or 50% or
0.5.
Example 6: Polygenic inheritance.
In wheat, kernel colour is determined by 2 gene pairs that produce colours ranging
from white to dark red. Dark red kernels are produced by plants with the genotype
AABB and white kernels are produced by plants with the genotype aabb. The more
alleles represented by capital letters in the genotype, the more red the kernels.
Thus, there are genotypes with between 4 and 0 capital letters and 5 classes of
colour. Let’s call the dark red kernels Class 1 and the white kernels Class 5.
What are the classes of offspring from a parent with the genotype AABb (Class 2)
and another with the genotype AaBb (Class 3)? In what proportions will they occur?
Steps 1 and 2:
Parent 1: gametes have either AB or Ab
Parent 2: gametes have AB, Ab, or aB
Step 3: Note that the Punnett Square has to increase in size and is not really a
square anymore!
AB
Ab
ab
AB
AABB
AABB
AaBb
Ab
AABb
AAbb
Aabb
Step 4:
The possible genotypes of the offspring are:
Genotype
Class of Kernel
Proportion
AABB
AABb
AAbb
AaBb
Aabb
Colour
1
2
3
3
4
2/6,
1/6,
1/6,
1/6,
1/6,
33%,
17%,
17%,
17%,
17%,
0.33
0.17
0.17
0.17
0.17
Example 7: Dihybrid cross without linkage.
In pea plants such as Mendel studied, a plant that is heterozygous for pod colour but
homozygous dominant for pod shape was crossed with a plant that is homozygous
recessive for pod colour and heterozygous for pod shape. What fraction of their
offspring will have a phenotype that is unlike either of the parent plants?
Recall that according to Mendel, the traits do not depend on each other and the
“factors”/genes are independently assorted. (Mendel’s Law of Independent
Assortment)
Step 1:
Let G represent green pod colour and g represent the allele for yellow pod colour.
Let I represent the allele for inflated pod shape and i represent the allele for
constricted pod shape.
Step 2:
Parent that is heterozygous for pod colour but homozygous dominant for pod shape:
GGIi (green pod, inflated)
Parent that is homozygous recessive for pod colour and heterozygous for pod shape:
ggIi (yellow pod, inflated)
Step 3: Note that if the parents are both heterozygous for both genes, the
Punnett square will have 4 types of gametes for each parent and thus 16
spaces for offspring genotypes. Some of the genotypes will end up being
the same though. Be careful counting up the genotypes and phenotypes.
gI
gi
Step 4:
GI
GgII
GgIi
gi
ggIi
ggii
The offspring that are ggii will have yellow, constricted pods. Thus ¼ of the offspring
will have a phenotype that is different from either of the parents. There are no
offspring that have green constricted pods.
Example 8: Dihybrid cross with linkage, no crossing over.
For Drosophila, the normal allele or wildtype allele is the dominant allele.
Use Figure 13. Write out all the genotypes and phenotypes of a cross between a
male that is heterozygous for leg length and heterozygous for eye colour (red/purple
alleles) and a female that is heterozygous for leg length and homozygous recessive
for eye colour. For this question, the male’s dominant alleles are on the same
chromosome.
Step 1:
Let L represent the allele for long legs and l represent the allele for short legs.
Let R represent the allele for red eyes and r represent the allele for purple eyes.
Step 2:
Male parent is LlRr and female parent is Llrr. However, since the genes are linked,
the notation looks like this:
Male:
L
R
l
r
Female:
L
r
l
r
Therefore, the male parent makes 2 types of gametes: LR and lr.
Therefore, the female parent makes 2 types of gametes: Lr and lr.
Step 3:
Lr
lr
LR
LLRr
LLRr
lr
Llrr
llrr
Step 4:
Possible Offspring Genotype
LLRr
LLRr
Llrr
llrr
Possible Offspring Phenotype
Long legs, red eyes
Long legs, red eyes
Long legs, purple eyes
Short legs, purple eyes
Question: Would the results be the same if the dominant and recessive alleles of the
male parent were on different chromosomes? Explain your answer using a Punnett
Square.
Example 9: Dihybrid cross with linkage and crossing over.
Use Figure 13. A male Drosophila that is heterozygous for aristae length and leg
length is crossed with a female Drosophila that is also heterozygous for both of
these genes. In the male, the dominant alleles are on the same chromosome and in
the female, the dominant alleles are on different chromosomes. In this case, the
chromosomes cross over during sperm formation.
What are the expected genotypes and phenotypes of the offspring?
Step 1:
Let A represent the allele for long aristae and a represent the allele for short aristae.
Let L represent the allele for long legs and l represent the allele for short legs.
Step 2:
The male parent is AaLl and the female parent is also AaLl. However, since the
genes are linked, the notation looks like this:
Male:
A
L
a
l
Female:
A
l
a
L
Since crossing over occurs in the male during meiosis, he will have a new
combination of alleles on some of his chromosomes and the chromosomes he will
put into his gametes will be any of the following four:
Male:
A
L
A
l
a
l
a
L
Step 3:
AL
Al
aL
al
Al
AALl
AAll
AaLl
Aall
aL
AaLL
AaLl
aaLL
aaLl
Step 4:
Possible Offspring Genotype
AALl
AAll
AaLL
AaLl (2)
Aall
aaLL
aaLL
Possible Offspring Phenotype
Long aristae, long legs
Long aristae, short legs
Long aristae, long legs
Long aristae, long legs
Long aristae, short legs
Short aristae, long legs
Short aristae, long legs
For more examples to work through, try this website:
http://www.schenectady.k12.ny.us/putman/biology/data/inheritance/intro.html
References:
1. http://www.nature.com/scitable/topicpage/dna-packaging-nucleosomes-andchromatin-310
2. http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/N/Nucleus.html#Nucl
eosomes
3. http://www.biochem.arizona.edu/classes/bioc462/462a/NOTES/Nucleic_Acids/
nucacid_structure.html
4. http://www2.estrellamountain.edu/faculty/farabee/biobk/BioBookgeninteract.h
tml#Polygenic%20inheritance
5. http://ghr.nlm.nih.gov/condition/hemophilia
6. http://staff.jccc.net/pdecell/evolution/polygen.html#polygen
7. http://waynesword.palomar.edu/lmexer5.htm
8. http://www2.estrellamountain.edu/faculty/farabee/biobk/BioBookgenintro.htm
l#The%20Monk%20and%20his%20peas
9. http://kentsimmons.uwinnipeg.ca/cm1504/geneticvariability.htm
10. http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/M/Meiosis.html
11.
http://www.biology.iupui.edu/biocourses/n100/2k2humancsomaldisorders.html