Agron 527
Homework 1
Due 2/1/02
Mendel probably failed to observed linkage between stem length (le) and pod shape (p or v)
for one of the following two reasons.
• Pod shape trait studied by him was governed by p, not by v that is linked to le.
• Mendel studied a small F2 population to study the di-hybrid segregation of le and v.
Assuming that the second scenario is correct, what are the possible explanations to justify
Mendel’s failure to detect the linkage between le and v loci?
Considerations to make:
1. The genetic distance between the two loci was 12 cM.
2. The mutation rates for both genes were zero.
==================================================================
Answer:
It is assumed that:
1. Mendel did study the v locus, which is linked to le.
2. The genetic distance between the loci is 12 cM.
3. Mutation rates for both genes were zero- no mutation.
Therefore, the genetic map distance between the loci studied by Mendel was 12 cM.
This means Mendel somehow managed to overlook the linkage between the two genes.
Allelomorphs of these two genes could be linked either in the coupling or in the repulsion
phase.
If these two genes were in the coupling phase linkage (LeV), then we will find the
following segregating ratio.
Recombinant gametes:
• There will be two classes of recombinant gametes.
o Lev
o leV
These two classes of gametes appear in equal proportion and total proportion of these two
classes of gametes is 0.12 (recombination fraction; 12 cM).
Thus, each gamete will appear = 0.12/2 = 0.06.
Parental gametes:
• Following two classes of parental gametes will be formed.
o LeV
o lev
The proportion of parental gametes = 1- 0.12 = 0.88.
Each of the two classes will appear at a frequency = 0.88/2 = 0.44
We will obtain the following genotypes from random mating of these four classes of gametes.
Coupling phase
LeV (0.44)
leV (0.06)
Lev (0.06)
lev (0.44)
LeV (0.44)
19.4
2.6
2.6
19.4
leV (0.06)
2.6
0.36
0.36
2.6
Lev (0.06)
2.6
0.36
0.36
2.6
lev (0.44)
19.4
2.6
2.6
19.4
Le-V-: 19.4 + 19.4 + 19.4 + 2.6 + 2.6 + 2.6 + 2.6 + 0.36 + 0.36 = 69.32
leleV-: 2.6 + 2.6 + 0.36 = 5.56
L-vv: 2.6 + 2.6 + 0.36 = 5.56
lelevv: 19.4
If these two genes were in the repulsion phase linkage, then we will find the following
segregating ratio.
Recombinant gametes:
• The two classes of recombinant gametes are:
o LeV
o lev
Each gamete class will appear = 0.12/2 = 0.06.
•
The frequencies for the following two parental classes will be = 0.88/2 = 0.44.
o Lev= 0.44
o leV = 0.44
We will obtain the following genotypes from random mating of these four classes of
gametes.
Repulsion phase
Lev (0.44)
leV (0.44)
LeV (0.06)
Lev (0.44)
19.4
19.4
2.6
leV (0.44)
19.4
19.4
2.6
LeV (0.06)
2.6
2.6
0.36
lev (0.06)
2.6
2.6
0.36
Le-V-: 19.4 + 19.4 +2.6 + 2.6 + 2.6+ 2.6 + 0.36 + 0.36 = 49.92
leleV-: 19.4 + 2.6 + 2.6 = 24.6
Le-vv: 19.4 + 2.6 + 2.6 = 24.6
lelevv: 0.36
lev (0.06)
2.6
2.6
0.36
0.36
The genotypes from coupling and repulsion phase linkages are tabulated along with those of
independent assortment in percentages.
Linkage
Le-V-
leleV-
Le-vv
lelevv
Repulsion
Coupling
Independent
Assortment
49.92
69.32
56.25
24.96
5.56
18.75
24.96
5.56
18.75
0.36
19.4
6.25
Ratios for 16 genotypes:
Linkage
Repulsion
Coupling
Independent
Assortment
Le-V-
leleV-
Le-vv
lelevv
8
11
9
4
1
3
4
1
3
0
3
1
Conclusions:
Linkage for two loci that are 12 cM apart in repulsion phase may escape the detection in an
F2 population if one studies a small segregating population. It is not impossible to get one
phenotype of the llvv type (0.36%) in a small population, since obtaining one is a matter of
chance. Probability of getting one llvv genotype is not zero. Thus, most likely le and v genes
in the cross, studied by Mendel were in the repulsion phase.
Homework 2
Due 6 PM, February 7, 2002
•
•
Resistance of soybean to the bacterial pathogen Pseudomonas syringae pv.
glycinea (Psg) is conferred by major resistance genes (Rpg). You are to study the
inheritance of resistance in soybean against this pathogen as a graduate student.
Resistance is dominant over susceptibility.
•
It is not clear whether the resistance of soybean to Psg race 1 is conferred by a
single or two Rpg genes.
•
If there are two genes responsible for resistance, then assume that at least a single
copy of each gene (wild type functional copy) is required to result resistance
against this race.
•
Assume that there is no linkage between the possible two Rpg genes.
•
What would be the minimum population size (F2) that you would like to
evaluate in order to determine the number of genes that confer resistance of
soybean against Psg race1?
Develop your hypotheses for the inheritance study after determining the minimum
number of F2 plants required in your investigation.
Answer:
•
•
•
•
It is known that resistance is dominant over susceptibility.
It is not known whether the resistance is governed by one or two genes.
If two genes govern the trait, then both are essential.
Both genes segregate independently- no linkage.
If a single gene governs the resistance we will observe 3:1 ratio.
If both genes are essential for resistance, then the 9:7 ratio will be observed. Because at
least one copy of each gene is essential for expression of resistance and genes are not
linked.
We need to determine the minimum population size (F2) for determining the number of
genes that confer resistance of soybean to Psg race 1.
Based on the information provided the resistance against Psg race 1 could be governed by
either one or two genes; and genotypic ratios depending upon the gene number could be
either 3:1 or 9:7:: resistance : susceptibility.
To determine the genetic ratio unambiguously we need study a population in which the
number of susceptible plants will be equally likely to have arisen from a 3:1 ratio or a 9:7
ratio. We can determine this ambiguous segregation ratio at which the observed ratio
resulting from either one or two genes produces equal χ 2 values at a probability = 0.05 as
follows.
The observed ratio at which hypothetical l1:1 and l2:1 ratios produce equal χ 2 values
is:
l1l 2 :1
Based on this general formula the ambiguous ratio for 3:1 and 9/7:1 is:
27
:1
7
= 1.964 : 1
The observed plants for dominant and susceptible classes among ‘n’ F2s can be calculated
as follows:
Susceptible =
1
n
2.964
Resistant
1964
.
n
2.964
=
The ambiguous segregation in terms of n is
n
1964
.
: n
2.964 2.964
This ratio is the borderline ratio that separates 3:1 from 9:7. By using the observed data
from this ratio and the χ 2 value (3.841)for this segregation at 1 df for  = 0.05, in the
Mather’s short formula for 2-class segregation of 3:1 we obtain the value of the minimum
family size of F2 (n) as follows:
1
χ2 =
(a1 − 3a2 )2
3n
.
n
n
− 2.3964
( 1964
)
2.964
3.841=
3n
2
.
n
( −21036
.964 )
=
3n
2
=
0122
. n2
= 94.3
3n
n = 94.31
Thus, we need to test 95 plants to be 97.5% (p=0.05, but we are only interested in
deviation in one direction from 3:1 towards more recessive in 9:7) sure that the
segregation we see is either 3:1 or 9:7.
The number of susceptible plants = 95 x 1/(2.964) = 32.
So if we see more than 32 recessives out of 95 plants, we assume 9:7 is correct and the
trait is governed by two genes. On the other hand if we see less than 32 susceptible
plants, then we will accept the 3:1 ratio or single gene inheritance.
The following hypotheses are developed for studying the inheritance of the Psg
resistance.
Hypothesis 0: The resistance of soybean against Psg race 1 is conferred by a single gene.
Hypothesis A: The resistance of soybean against Psg race 1 is conferred by two genes.
Homework#3
Following are two summaries with minor editing.
===============================================================
Comparative Genome Organization in plants: From Sequence and Markers to
Chromatin and Chromosomes
Summary
Introduction:
Comparative studies have shown that various biological structures and functions are
conserved among the living organisms. These have been proved by cytological and
molecular studies.
Molecular studies have shown that structures like ribosomes, ribozymes and features of
genetic code are conserved across the living organisms. Such studies provided useful
markers for evolutionary studies. Findings from comparative studies have encouraged the
biologist to determine the whole genome sequence. It is believed that knowledge of the
whole sequence of an organism will aid in the isolation of sequences common in other
related organisms, and thus help in isolation of genes in related species.
The Linear DNA Sequence:
Sequencing projects have shown that the double-helical structure of DNA and its
composition (A, T, G, C bases) is universal in nature. It has also been shown that
chromosomes start and end with telomeres. In the plant kingdom, Arabidopsis was
chosen for the sequencing project mainly because it has a small genome of 130-140Mbp
that is diploid with 5 pairs of chromosomes.
Genome size and plant niche are significantly correlated, but there isn’t any clear
correlations between chromosome number and any plant characteristic except for
polyploidy.
Sequencing is followed by annotation and identification of genes. Arabidopsis has
approximately 25,000 genes that represent most of the genes found in plants with much
larger genomes. Then why is there such a discrepancy in the sizes of the genome? It is
mainly because of the difference in the number of repeat sequences from one species to
another. About 50-90% of the genome of higher eukaryotes is composed of these DNA
motifs. Though genomic sequence can tell us about the nature of genes and their
function, sequencing doesn’t distinguish between modified and unmodified bases, and
fails to tell us about chromatin packaging and three-dimensional organization of the
chromosomes.
Repetitive DNA Sequence and the Large-Scale Organization of the chromosome:
Before genomes of different organisms can be compared, the length of the sequence gaps
must be determined, the homogeneity of repeat motifs should be known, and the extent of
variation within the motifs should be known in order to ascertain the function of the
repeat elements in the genome. Some of the sequence repeats have been highly
conserved from one species to another like the rDNA genes, but some repeats are highly
variable even between accessions of a species. The study of repetitive DNA sequence
motifs and their chromosomal distribution has considerable potential for understanding
genome evolution and sequence components. It was discovered that amidst repetitive
sequences, especially in the centromeric region, lie some genes. The concept of families
of repeat sequences has been developed to understand these regions of the genome.
Classes consist of 1) tandem repeats 2) retroelements and 3) telomeric sequence.
Cytogenetic methods have used in situ hybridization methods to determine the
localization of the repeat sequences.
rDNA:
The DNA that codes for the rRNA is known as rDNA. It is highly conserved and
consists of tandem array of repeating units of the rRNA genes encoding 18S, 5.8S and
26S rRNAs and spacers (transcribed and non-transcribed) of approximately 10 Kb in
plants. Repeat units and the 5s rRNA genes are localized at specific regions of the
chromosomes which makes them usable as markers. Evolutionary trends have been
studied due to the general correlation of speciation rates to changes in chromosomal
distribution of these repeat units. rDNA repeats represent ~10% of the genome.
Telomeres:
They are the specialized structures present at the end and start of the chromosomes. They
are highly conserved regions of hundreds of tandem repeats with the sequence similar to
TTTAGGG. The enzyme telomerase is required for telomeric replication. The enzyme
supplies an RNA template for telomere replication. Telomerase enables chromosomal
stabilization and repair.
Centromeres:
They are the attachment site of microtubules during cell division. Centromeres are often
composed of tandem repeats, which are highly conserved and are defined cytologically
by primary constriction. Centromere-associated repeats represent a considerable
percentage of the genomic DNA. Despite analysis of the structure and proteins
associated with the centromere, comprehensive information about centromeric DNA
sequence is lacking. Some scientists are of the opinion that the tandem repeats play a key
role in centromere function and chromosome segregation. Recent analyses of
centromeres have shown that it as not devoid of genes as was previously believed. A few
genes and a wide range of vestigial and presumably inactive mobile elements have been
identified. The centromere consists of a central, repetitive core, flanked by moderately
repetitive DNA that has few recombination and then by regions with mobile elements and
normal recombination rates.
Transposable Elements and Retroelements:
They are discrete components of the plant genome that replicate and reinsert at multiple
sites by a complex process. Depending on the method of excision and reintegration,
these mobile elements are classified as either Type I, which uses an RNA intermediate,
e.g. retrotransposon, or Type II, those existing exclusively as DNA.
Retroelements are very heterogeneous and found in the whole of the plant kingdom
indicating its ancient nature. It is hypothesized that retroelements are more around the
centromere region so as to limit the disruption of genes. Retroelements are a source of
biodiversity as they can cause a mutation when present in a gene. It is estimated that
80% of mutations detected in Drosophila are caused by retrotransposons. Transposons
can partially or completely restore gene function and may even create new gene
functions, thereby contributing to evolution. It has been shown that stress activates
retroelements. The sequence of degenerate and potentially active retroelements gives
valuable information about genome evolution and phylogenetic relationships.
Retroelement amplification leads to large genomes and loss can occur in a specific
manner leading to species-specific composition of retroelements.
Simple Sequence Repeats (SSRs):
SSRs or microsatellites are small nucleotide repeats (~ up to 5bp) that are present all
along the eukaryotic genome. They provide highly informative and polymorphic markers
for plant, fungal and animal fingerprinting.
Tandem Arrays of Repetitive DNA:
They can provide useful markers for chromosome identification, and their presence and
distribution can provide evidence for evolutionary changes. Evidence does not exist for a
constant mutation rate. Rather, bursts or evolutionary waves of mutations occurred.
Tandem arrays are usual transcription silent.
DNA Sequence in the Chromosome:
The packing of the genomic DNA can directly affect aspects of RNA transcription, DNA
replication, recombination, DNA repair, and chromosome segregation.
Methylation:
Cytosine methylated DNA is extensive. It is an important gene regulating mechanism.
Reports have correlated some Methylation patterns to reduce levels of gene expression,
whereas other patterns are correlated to normal regulation of developmentally important
genes. Probably, methylation occurs at symmetrical sites in the DNA molecule, in
animals, whereas in plants, methylation does not necessarily occur at symmetrical sites.
Usually, DNA methylation is a terminal stage of differentiation, but changes in patterns
have been noticed during plant development (e.g. meiosis and embryogenesis). DNA
methylation also helps in maintaining the chromosome stability. The DNA
methyltransferases are known to participate in DNA repair and stabilize nucleoprotein
assemblies required in the inactivation and imprinting of chromosomes.
Structure and Packaging of Linear DNA into Chromosomes:
The DNA is wrapped around the basic proteins called histones forming nucleosomes
connected by linker DNA. Repetitive sequences probably play a key role in stabilizing
this structure.
Chromatin Remodeling and Histone Acetylation:
Histone acetylation is known to change the structure of the chromatin. It does it by
modulating the position of nucleosomes. Changes in nucleosome position affect the rate
of transcription by blocking the access of transcriptional factors to the promoters.
Remodeling may be a requirement for replication of condensed, inactive regions of the
genome.
The Three-Dimensional Nucleus
Genome Architecture:
Two-dimensional linear models inadequately explain gene regulation, thus architecture is
important in understanding gene regulation. Architecture refers to the genomes threedimensional structural organization within the nucleus and extends to the dynamics and
relationships between structure and function. The genome architecture is of prime
importance as the functional regulation of DNA behavior depends on genome
organization. DNA packing and unpacking, replication, repair, mutation and
transcription are tissue specific and depend on the dynamic architecture of genome
organization.
Packaging of Nuclear DNA:
It is believed that an intranuclear framework provides a functional organization for the
genome. But the existence of a nuclear matrix or chromosomal skeleton remains
controversial. Most of the major cytoskeletal proteins like actin and tubulin have been
found in the nucleus but their exact function and significance are not known. “The
higher-order structure of the chromatin fiber and the organization of chromatin domains
in the nucleus appear to have a profound influence on gene expression.” Within the
nucleus there is compartmentalization of individual chromosomes, euchromatic and
heterochromatic regions, and the nucleolus. The active genes tend to move to the
periphery near the nuclear membrane where RNA transcripts are formed. Nucleoli, the
sites of rRNA synthesis, are spherical compartments within the nucleus with no defined
boundaries; they move and fuse during interphase of the cell cycle.
Genomics, Chromosomes, Evolution, and the Nucleus:
The chromosome, chromosome segment, gene, and DNA sequence are levels of genome
evolution that plant breeders aim to control and direct. The plasticity of the genome and
rapid amplification and fixation of advantageous novelties has been shown. Organization
of the chromosome has a fundamental influence on these evolutionary processes. It is
now known that tandem repeats in relation to chromosome structure are present at
Telomeres and Centromeres and also that retroelements represent about 50% of the DNA
in the genome. Comparative analysis has been useful in understanding genome
organization.
Thus from the studies of the genome organization and gene functions much can be known
about the fundamental processes, such as chromosome pairing, segregation, gene
organization and expression, and its direct implication on the aims of biologists. The
structure and some sequence of the DNA and its organization have been conserved in all
the organisms showing us the importance of these in the maintaining the life cycle.
SUMMARY #2
Comparative Genome Organization in Plants: From Sequence and Markers to
Chromatin and Chromosomes
I.
The Linear DNA Sequence
Complete sequencing of various plant and animal genomes has affirmed the
following: 1) the double helical DNA model is continuous from telomere to
telomere, and 2) no other bases beside adenine, guanine, thymine, and cytosine are
present.
Arabidopsis was the first plant genome to be completely sequenced because of
its small size (130-140Mbp). Its genome contained nearly 25,000 genes. However,
the small size of the genome does not correlate to a smaller number of genes. In fact,
most of the genes found in plant species with much larger genomes (pines have
~23,000Mbp) were also found in Arabidopsis. This is because the amount of highly
repetitive sequences (non coding regions) found in Arabidopsis was much less than
what is found in the genomes of larger plants.
II.
Repetitive DNA Sequences and the Large-Scale Organization of the
Chromosome
Most of the DNA code in humans and Arabidopsis that has yet to be sequenced
lies within regions of highly repeated DNA. These repeated stretches make it hard
for accurate reconstruction of the genetic sequence since they are over several
hundred base pairs long and fall outside the limits of typical sequencing methods.
Therefore, the logical ordering of sequenced contigs is very difficult due to the high
degree of repetition. It is interesting to note that some genes have been discovered in
the highly repetitive regions around the centromere. However, there is no data to
support that these repeated sequences play a role in gene expression or whether they
simply contribute to the overall size of the genome.
Many of the repeated sequences are highly conserved in all eukaryotic
species. Conversely, some repeated sequences show a great deal of variation and can
be different within two individuals of the same species. Therefore, the study of
differences between these highly conserved sequences between species can offer
information on the evolution of their respective genomes. Some more common
classes of repeated sequences are tandem repeats, retroelements, and telomeric or
rDNA units.
In situ hybridization can be used to provide information about the location of
these repeated sequences along the chromosome. Using denatured chromosomes and
labeled probes complementary to the repeated sequence in question, hybridization can
be conducted on the surface of a slide and the pairing can be visualized with a
microscope. The article shows some very nice pictures of this procedure that provide
information on where certain repeated sequences can be found on the chromosomes.
Various studies using in situ hybridization have been able to show where the majority
of these repeat groups occur on chromosomes. Other advantages of this technique are
its ability to scan for DNA from viruses and mitochondria without prior sequence
knowledge of the flanking regions.
III.
R-DNA
The r-DNA subunits are repeated hundreds or thousands of times and may make
up as much as 10% of the genome. The 45S rDNA loci consist of the 18S, 5.8S, and
26S rRNA genes (the S nomenclature is based on the unit’s sedimentation rate in a
centrifuge) that are reach approximately 10kb long in plants. These units assume
specific positions on chromosomes and can be subsequently used as markers for
chromosome identification. Due to the high degree of conservation of these
sequences, probes that have been isolated from one species can be used to identify the
similar unit in most other eukaryotic species.
Looking at changes in the chromosomal distributions of these subunits can
provide incite about rates of speciation and evolutionary trends.
IV.
Telomeres and Centromeres
Telomeres are highly conserved regions of DNA near the ends of
chromosomes. It is thought that conservation of this region is vital to the DNA
replication process. The region is constructed from a highly repeated short DNA
sequence (TTTAGGG). It is very species specific, but these repeats can be hundreds
of units long in many eukaryotes. Even within species, the number of telomeric
repeats can be different from one chromosome to another. Telomeres require a
special telomerase enzyme to aid in their replication. This telomerase not only
provides the RNA template for telomere replication, but it also serves as a substrate to
stabilize and repair damaged chromosomes.
Centromeres are also very highly conserved regions of DNA within the
chromosome. Centromeres provide the location for proper spindle attachment during
mitosis and meiosis. Therefore, it is vital for this region to remain conserved for
proper cell replication and gamete formation. It is believed that the centromeres are
made up in large part by a repeated 180bp sequence. This repeat is very prevalent
and can account for a large portion of the total genomic DNA of a given species
(0.3% in humans and 3% in Arabidopsis). Sequencing within the centromeric region
of Arabidopsis has revealed a large number of inactive mobile elements. One study
has been conducted to reveal the DNA sequences responsible for proper centromere
function. It was found that centromeres contain a central repetitive core, flanked by
moderately repetitive DNA with a low recombination rate, which is then flanked by
regions of mobile elements with normal rates of recombination. However, it is not
believed that the repeats alone are responsible for centromere function since some of
them are even more abundant in DNA outside of the centromere.
V.
Transposable Elements and Retroelements
Class I transposable elements (retroelements) are pieces of DNA which can
replicate and move from one region of the genome to another. These elements can
makeup half of the genome’s nuclear DNA and usually include 2-3 open reading
frames (ORFs) and are nearly 5kb in length. Depending on the species, transposable
elements can be found pretty much anywhere in the genome. In general, they are not
as prevalent in the rDNA and centromeric regions of the genome where there is a lot
of tandemly repeated DNA sequences. However, in Arabidopsis there is a high
proportion of transposons in the centromeric region of the chromosomes, where a
very few genes are located for disruption by retrotransposons.
Retroelements also play an active role in gene regulation and mutation.
Depending on where they insert themselves in a specific gene sequence, the
expression of that gene may be mutated or knocked out completely. In Drosophila,
retroelements are believed to account for nearly 80% of the mutations uncovered.
Retroelements may also serve in species evolution by altering gene functions. For
example, a retroelement may carry a promoter which, when inserted, activates a gene
that was previously unexpressed. Similarly, the element may alter the function of an
existing gene that leads to varied expression.
VI.
Simple Sequence Repeats
SSRs are series of repeated sequences. Allelic discrimination can be achieved if
the number of SSRs at a given site is different between alleles of a given gene.
Primers that flank both sides of the SSR can be synthesized for PCR amplification of
the repeated region. Variation can be visualized through agarose gel separation and
ethidium bromide staining.
Similarity within SSR sites of related species can provide a tool for estimating the
genetic distance between the two. This type of information is very valuable to those
studying the evolution of plant and animal species.
VII.
Methylation
DNA methylation is a process in which cytosine residues, of CG doublets in
particular, are methylated in the 5-position of the cytidine ring. Heavily methylated
regions of DNA are usually transcriptionally inactive. Methylation near genes or
gene promoter regions has resulted in decreased levels of gene expression. Therefore,
methylation has been shown to be an effective method of gene regulation.
Methylation occurs following DNA replication on the newly synthesized
DNA strand at specific nucleotide sequence sites. Methylation patterns are copied
from the parent DNA strand using maintenance methylases. However, in plants, the
methylation process does not seem to mimic that of the parent strand. Instead,
methylation appears to be conducted completely anew with no regard to the parent
strand. Studies have also shown that the amount of methylation appears to decrease
after repeated cycles of DNA replication. This results in regions of hemimethylated
DNA and ultimately in regions of unmethylated DNA. Different DNA methylase
enzymes have also been found which could be tissue specific or specific to a pattern
of methylation (symmetrical or asymmetrical). A demthylase enzyme has been
discovered which may have implications as a new mechanism for gene
activation/regulation.
VIII. Structure and Packaging of Linear DNA into Chromosomes
The higher order packaging of inactive DNA in the nucleosomes is not a static
system. This organization is found to be highly variable in both plants and animals.
In fact, in two closely related species such as rye and wheat, significant variation has
been seen in the length of linker DNA between telomeres and expected repetitive
DNA sequences in the chromatin material. These differences were uncovered using
micrococcal nuclease (MNase) to digest the linker DNA that connects nucleosome
particles together. In fact the sheer difference in sensitivity to MNase between these
closely related species also denotes a difference in DNA packaging.
Repetitive DNA sequences such as tandem arrays show a characteristic
arrangement around the nucleosomes and most likely play a stabilization role in DNA
packaging and chromatin condensation. Repeated adenine sequences of 4-6 bases in
phase with the helix promote bending that subsequently aids in nucleosome assembly.
Any DNA remaining after the repeats have been packaged can easily be fit into the
nucleosome.
IX.
Chromatin Remodeling and Histone Acetylation
Histone acetylation and chromatin remodeling play a similar role in gene
regulation as the DNA methylation of cytosine residues. Expression of gene products
may be altered or silenced depending on the extent of remodeling and acetylation. It
has been proposed that regions in the human genome where chromatin remodeling
occurs at a high frequency between mitosis and meiosis could be hot spots for
recombination.
X.
Packaging of Nuclear DNA
In contrast to earlier claims that the nucleus was an unorganized mess, there is a
preponderance of evidence to prove that the nucleus is actually highly organized and
functional. It has been shown that centromeres and telomeres occupy specific parts of
the nucleus in many species. Furthermore, the nuclear material can be dissected into
unique parts including chromosomes, euchromatic and heterochromatic regions, the
nucleolus, and regions of active DNA synthesis. However, the presence of nuclear
matrices, like nuclear scaffolds, cages, and compartments is argumentative. Most of
these documented structures have appeared in laboratory situations that do not reflect
cellular conditions in vivo.
The haploid chromosome consists of unique transcription subunits that
assume a specific formation that only the homologous chromosome could recognize
and pair with during meiosis. This “best fit” mechanism ensures proper alignment of
all homologous chromosomes.
The strongest evidence for nuclear compartmentalization comes from the
nucleoli. These are spherical units of the nucleus that have no boundaries. The
composition of nucleoli is very different from other nuclear components and
depending on cell type, may assume different locations within the nucleus.
XI.
Genomics, Chromosome, Evolution, and the Nucleus
The current depth of understanding concerning the nucleus and
chromosomes is far greater than it was only a few years ago. This will aid biologists
in developing new methods for gene cloning, evolutionary studies, gene transfer, and
gene expression. Future plant breeding efforts can focus on controlling plant
evolution in a way that conserves both biodiversity and the environment while still
introducing new traits of agronomic importance. Studies have shown that
advantageous genetic traits quickly move towards fixation within a species.
Increased understanding of the processes that lead to genetic variation will provide
more avenues for researchers to use. Furthermore, understanding the evolutionary
processes that have already occurred within the plant genome will enable breeders to
understand the consequences of any future changes. Finally, new technologies such
as in silico analyses, high-throughput DNA and protein studies, digital imaging
methods, and in vivo methods for studying the dynamic chromosome structures and
interactions will make sure that our understanding of the genome continues to
increase.
Homework #4
Answer to Q#1:
It appears from the results that two genes are involved in the development of red flower
color in the crosses #1 aand #2, because of the 9:7 ratio. In the cross #3 only one gene is
mutated to result white segregants and a 3:1 ratio.
Assuming that the same two genes are segregated in crosses 1 and 2, and one of these is
mutated in parent 6 we can obtain the genotypes of the parents as follows.
•
A and B are dominant alleles that encode functional enzymes necessary for red
pigment synthesis, whereas a and b are recessive or mutant alleles that encode
non-functional enzymes for pigment biosynthesis.
•
In P2 (aabb) both gene are mutated, while P1 (AABB) carries the wild type
(original functional type) genes. Then only F2 ratio will be 9:7::red:white (shown
below in Table 1- not necessary to show this table of genotypes in exam).
•
P3 (aaBB) carries one mutant (aa) and one wildtype gene (BB), whereas P4
(AAbb) carries mutant allele for B gene and wildtype A gene. Therefore, both
parents produce white flower. But when they were crossed, F1 produced red
flower due to complementation. Enzyme produced by A from P4 and enzyme B
produced by P3 can complete the pathway and synthesize the red pigment.
•
P5 (AABB) carries wildtype alleles in both A and B loci. But P6 (Aabb) carries a
mutation in the B gene, and therefore, it produces white flower.
Table 1: F2 plants with at least one copy of both wildtype genes shown by bold letters.
AB
Ab
aB
ab
AB
AABB
AABb
AaBB
AaBb
Ab
AABb
AAbb
AaBb
Aabb
aB
AaBB
AaBb
aaBB
aaBb
ab
AaBb
Aabb
aaBb
aabb
Summary of parents’ genotypes
Parent
P1
P2
P3
P4
P5
P6
Genotype
AABB
aabb
AAbb
aaBB
AABB
AAbb
Color
Red
White
White
White
Red
White
Models to explain the flower color development
The red pigment is the end product of a metabolic pathway. Several enzymes are
involved in the synthesis of the red pigment. Each enzyme encodes an enzyme of the
pathway. For example, A and B genes encode enzyme A and B, respectively. Both
enzymes are essential for the synthesis of the red pigment. Enzyme A converts a
precursor to the product A, whereas enzyme B converts the compound A to B.
In cross #1 between P1 (AABB) and P2 (aabb) P1 contributed both genes, whereas P2
had none. But in F1 one copy of A and B genes produced both enzymes, and therefore,
flower color was red. In F2, progeny carrying both wildtype genes A and B (bold lettergenotypes in Table 1) were with red flowers, while absence of any of these two wildtype
genes were carrying white flowers.
We therefore, draw a simple model to explain the observation.
Precaursor
Precaursor
Red pigment
Compound A
Enzyme A
Compound B
Enzyme B
Red pigment
In cross #2 wildtype A gene contributed by P4 and B is by P3. Again the above model
should explain the results.
In cross #3 only B gene is mutated in one of the parents. The A gene is wildtype in both
parents. Therefore, we show only 3:1 ratio, not 9:7. Again same model should explain the
observation of this cross. In this case only enzyme B is missing in P6.
Answer to Q#2.1:
Let R be the dominant and r the recessive allele for seed shape and let A be the dominant
and a the recessive allele for flower color.
P1 (RRAA) x P2 (rraa) ⇒ F1 (RrAa) with the alleles in coupling phase.
F2
Genotype
Segregation
R-AR-aa
rrArraa
Count
450
80
75
115
Variable
a
b
c
d
The quick test (ad/bc) = [(450*115)/(80*75)] = 8.625.
8.625 > 1, thus we suspect linkage for coupling phase.
H0: Expected ratio 9:3:3:1.
Genotype Observed Expected
X2
R-A450
405
5
R-aa
80
135
22.407
rrA75
135
26.667
rraa
115
45
108.889
Total
162.963
A χ2 with 3 degrees of freedom at 0.01 p-value is 11.3. 162.963 > 11.3. Thus, we reject
the H0: 9:3:3:1. Now we check to see if the deviation is due to distorted segregation of
individual traits or due to linkage between the loci.
For R-:rr, the H0: 3:1.
Genotype Observed Expected
R530
540
rr
190
180
Total
X2
0.185
0.556
0.741
A χ2 with 1 degrees of freedom at 0.01 p-value is 6.63. 0.741 < 6.63. Thus, we accept
the H0: 3:1. There is no evidence of a distorted segregation for the seed shape trait.
Similarily, for A-:aa, the H0: 3:1.
Genotype Observed Expected
A525
540
aa
195
180
Total
X2
0.417
1.25
1.667
A χ2 with 1 degrees of freedom at 0.01 p-value is 6.63. 1.667 < 6.63. Thus, we accept
the H0: 3:1. There is no evidence of a distorted segregation for the flower color either.
The final test is H0: No linkage between loci.
χ2 = 162.963 – 0.741 – 1.667 = 160.555 with 1 df.
160.555 >> 6.63. Thus we reject the H0: No linkage between loci. Therefore, there is
linkage between the two traits.
Answer Q#2.2.
P3 (RRaa) x P4 (rrAA) ⇒ F1 (RrAa) with the alleles in repulsion phase.
F2
Genotype
Segregation
R-AR-aa
rrArraa
Count
450
220
235
5
Phenotypic
classes
a
b
c
d
The quick test (ad/bc) = [(450*5)/(220*235)] ≈ 0.044.
0.044< 1, thus we suspect linkage for repulsion phase.
H0: Expected ratio 9:3:3:1.
Genotype Observed Expected
R-A450
511.875
R-aa
220
170.625
rrA235
170.625
rraa
5
56.875
Total
X2
7.479
14.288
24.289
47.315
93.37
A χ2 with 3 degrees of freedom at 0.01 p-value is 11.3. 93.37 > 11.3. Thus, we reject
the H0: 9:3:3:1.
Now we check to see if the deviation of the ratio from the 9:3:3:1 ratio is due distorted
segregation or due to linkage between the loci.
For R-:rr, the H0: 3:1.
Genotype Observed Expected
R670
682.5
rr
240
227.5
Total
X2
0.229
0.687
0.916
A χ2 with 1 degrees of freedom at 0.01 p-value is 6.63. 0.916 < 6.63. Thus, we accept
the H0: 3:1. There is no evidence of distorted segregation for the seed shape trait.
Similarly, for A-:aa, the H0: 3:1.
Genotype Observed Expected
A685
682.5
aa
225
227.5
Total
X2
0.009
0.027
0.037
A χ2 with 1 degrees of freedom at 0.01 p-value is 6.63. 0.037 < 6.63. Thus, we accept
the H0: 3:1. There is no evidence of distorted segregation for the flower color trait.
The final test is H0: No linkage between loci.
χ2 = 93.37 – 0.916 – 0.027 = 92.417 with 1 df.
92.417 > 6.63. Thus we reject the H0: No linkage between loci. Therefore, there is
convincing evidence of linkage between seed shape and flower color for this data.
Answer Q#3. A test for independence can be used to determine if there is any linkage
between plant height and nematode resistance.
H0: No linkage between plant height and nematode resistance.
Observed:
Class
Resistant Susceptible
Tall
550
100
Dwarf
80
150
Total
630
250
Ratio: Res:Sus 0.716
0.284
Total
650
230
880
Ratio: Tall: Dwarf
0.739
0.261
Expected:
Class
Tall
Dwarf
Total
Resistant Susceptible
465.34
184.6559
164.659
65.431
630
250
Total
650
230
880
In all cases the (deviation)2 = 7167.62.
χ2:
Class
Tall
Dwarf
Total
Resistant Susceptible
15.402
38.813
43.527
109.689
58.929
148.502
Total
54.215
153.216
207.431
Total χ2 = 207.431 with 1 degree of freedom. 207.431 > 6.63, thus we reject the H0: No
linkage. Note that this method does not account for associations due to factors other than
linkage. Further experimentation is needed since the mode of inheritance is not known.