Download Genetics 314 Spring, 2004

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Genetics 314 - Spring, 2004
Exam 3 – 100 points
1. You are working for a company that is investigating the use of chemicals to
control various functions/activities in the cell. The first activity in the cell is the
company would like you to target is chromosome formation. They suggest
chemicals that prevent the synthesis of either a) basic proteins found in the
nucleus or b) acidic proteins found in the nucleus. Would either or both
chemicals interfere in chromosome formation? If yes explain how.
a) basic proteins
Yes it would interfere with chromosome formation.
If basic proteins found in the nucleus were not synthesized it would have a
major affect on the histone proteins necessary for chromatin organization
and chromosome condensation. The histone proteins 2A, 2B, 3 and 4 (2
copies each) form an octomer called the nucleosome. DNA will wrap around
the nucleosome initiating the first step of chromatin organization and
condensation. Histone 1 is needed to hold the nucleosomes together in the
solenoid structure so the loss of histone 1 would prevent the second step of
chromosome organization/condensation.
b) acidic proteins
Yes it would interfere with chromosome formation.
The acidic proteins in the nucleus are involved in forming the looped
domains of chromatin during interphase and help form the acid protein
scaffolding necessary for the final step of chromosome condensation.
Without these proteins chromatin organization would be affected, possibly
affecting proper gene transcription and chromosome condensation would not
occur if the acid protein scaffolding could not form.
2. Your company now wants you to work on controlling the cell cycle. In one case
they want you to stop the cell cycle in interphase and the other they want you to
keep the cell in mitosis. For both cases name two ways to achieve your
company’s goals and briefly describe why your approaches would work.
a) stop in interphase
To stop or keep a cell in interphase you could target several components in
the cell cycle. MPF formation could be prevented by preventing the
synthesis of cyclin or cyclin binding to cdc2. A second way to keep the cell in
interphase is to prevent the activation of MPF. Without activated MPF at
the necessary concentration in the cell, cell division will not begin and the cell
will stay in interphase.
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b) stop in mitosis
There are two ways to halt or keep a cell in mitosis. The first is to prevent
the deactivation of MPF. This can be accomplished by preventing the
degradation of cyclin or by adding cyclin to the system to keep the level of
cyclin high in the cell. If decrease in activated MPF is what signals the end of
mitosis so keeping the activated MPF level high will hold the cell in mitosis.
A second method would be to interfere with microtubule formation. The
microtubules are needed to align the chromatids on the metaphase plate and
separate the daughter chromosomes. If the chromatids do not have the
proper microtubule tension the centromeres will not separate and the cell
will stay in the metaphase stage of mitosis.
3. What are three differences between mitosis and meiosis that affect
recombination?
1. The chromosome number is reduced by half in meiosis while it remains
constant in mitosis.
2. Homologous chromosomes pair in meiosis but do not pair in mitosis.
3. Crossing-over can occur between non-sister chromatids in meiosis but not
in mitosis.
4. You discover a weed that once it invades a region quickly replaces the native
vegetation. When you examine the plant you find it releases a chemical that
prevents post-meiotic mitotic divisions in other plants.
a) Could this explain how it rapidly takes over a region? Briefly explain your
answer.
Yes, post-meiotic mitotic divisions are needed in plants immediately after
meiosis to produce the needed nuclei for double fertilization. Without these
divisions the gametes would not be functional and no seed would develop on
the native plants. This would allow the weed to rapidly spread through out
the area.
b) If animals were exposed to this chemical could it affect them like it does the
plants? Briefly explain your answer.
If all the chemical did was only inhibit post-meiotic mitotic divisions during
gamete formation then the answer would be no. Unlike plants, after meiosis
in animals the gametes contain all the nuclei needed for successful
fertilization. So inhibiting post-meiotic mitotic divisions would not affect
animals like it does plants.
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5. You are creating a gene and want to get it into the DNA of a plant cell. You
decide to use a transposable element to insert you gene.
a. Why would a transposable element be a good choice to get a gene into a plant’s
DNA?
Transposable elements are self-inserting so if the element could be delivered
into the nucleus there would be an excellent chance for it to be incorporated
into the cell’s DNA.
b. What could some of the problems be with using a transposable element to move a
gene into a plant?
The problems relate to the random insertion sites of the transposable element
and the unstable nature of the element. The insertion site for the element can
not be targeted so it is possible that the element could insert in a critical gene
shutting down expression of that gene. The element could also move which
could affect the introduced gene’s expression depending on where it inserts
(such as into constitutive chromatin). If the element used was a replicative
transposable element it could duplicate itself and inserting into other regions
of the genome. This could knock out additional genes and increase the level
of expression of the introduced gene by increasing the number of copies of
the gene in the genome.
6. You are studying some birds and find something interesting. You sex all the birds
in an isolated colony and find they are all female. You come back later and there
are still only adult females but also eggs and baby chicks that are all males.
Assuming no visiting males:
a) Can this be explained genetically? Briefly explain your answer.
Yes, in birds the female is the heterogametic sex and the male is the
homogametic sex. Because the female is the heterogametic sex, if an
unreduced gamete is produced during the equational division of meiosis or
possibly fertilization of by the polar body from the equational division occurs
it is possible to recover a 2N gamete with two X chromosomes. This cell
would have all the genetic information for a male chick. Hence without a
male bird for fertilization male chicks could be produced from the females.
b) Could this phenomenon be used to increase the population number of endangered
animal species? Why or why not?
In birds possibly since the males could be mated back to the females but
there would be a definite decrease in genetic diversity. In non-avian species
such as mammals the answer would be no, it could not be used to increase
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population number. In mammals the female is the homogametic sex so it
could only produce females from an unreduced gamete.
7. You find a plant that has 66 chromosomes. Based on chromosome pairing in
meiosis you determine the plant is a allohexaploid.
a) What is the somatic (2N) chromosome number of the plant?
2N = 66
b) What is the gametic (N) chromosome number of the plant?
N = 66/2 = 33
c) What is the monoploid (X) chromosome number of the plant?
X = 66/6 = 11
d) What does it mean that the plant is an allohexaploid in terms of the origin of the
chromosome sets?
If a plant is an allohexaploid it means the chromosome sets came from two or
more related species.
8. You cross two plants, one a tetraploid (4X) and the other a diploid (2X). You
discover the progeny grow more vigorously than the diploid parent and produces
large fruit. You think you have a new major crop that you can make money on by
selling seed from your progeny.
a) Why does the plant grow faster and have bigger fruit than its diploid parent?
The plant is a triploid (3X) plant so it contains an extra set of chromosomes.
The set of genes associated with the extra set of chromosomes allows for
increased expression of some traits. In addition, since a 3X plant would be
sterile more energy is spent on cell growth and development. The
combination of the extra genes and no gamete formation result in a more
vigorous plant with larger fruit.
b) Will your plan to get rich from selling seed from your new plant be successful?
Why or why not?
You will not get rich from selling seed from the 3X plant because the plant
will be sterile. It is not possible to divide 3 chromosome sets equally so the
gametes will get an unbalanced number of chromosomes rendering then nonviable.
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9. You find another plant that seems to also have a chromosome change. While
there is no change in the chromosome number and no change in the length of the
chromosomes there appears to be a problem in meiosis because the plants are
partially sterile.
a) What chromosome change could cause the plants to be partially sterile?
If there is no change in chromosome number and no change in chromosome
lengths the chromosome change that could lead to partial sterility would be
an inversion. With inversions there is no change in chromosome length, just
a reversal of the gene order for a section of the chromosome.
b) What conditions need to be present for problems in gamete formation to occur?
For the plant to be partially sterile due to an inversion the plant must be
heterozygous for the inversion. This means it has one normal chromosome
and one inverted chromosome. Only being heterozygous for the inversion
can a loop structure be observed during meiosis, which indicates the presence
of an inversion. In addition the duplications and deletions that lead to the
partial sterility will not occur unless there is a cross-over within the inverted
region.
10. You discover two cases of Down’s syndrome, one caused by aneuploidy and the
other caused by translocation.
a) Cytogenetically how do they differ?
For Down’s due to aneuploidy the condition is due to random nondisjunction of chromosome 21 during meiosis in one of the parents producing
a N+1 gamete. When combined with a normal N gamete it produces an
individual that is 2N+1. In humans this individual would have a somatic
(2N) chromosome number of 47 while both parents would have a somatic
(2N) chromosome number of 46.
Down’s due to a translocation is due to a translocation of a piece of
chromosome 21 onto chromosome 14 in one of the parents. This is a
permanent change in that parents chromosomes but due to still having only
two doses of chromosome 21 (one normal chromosome 21 and the
chromosome 21 attached to chromosome 14) the individual appears normal.
Such an individual is called a carrier. A Down’s child occurs if during
chromosome segregation during meiosis the normal 21 and the translocated
chromosome end up in the same gamete. If this gamete combines with a
normal gamete then the individual would have 3 doses of chromosome 21 (2
normal chromosome 21s and the translocated piece of chromosome 21). The
presence of 3 doses of chromosome 21 is what leads to Down’s syndrome.
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Cytogenetically a translocation would be observed in both the child and one
of the parents.
b) Why is it important to know the difference in the cause of Down’s syndrome?
Down’s due to aneuploidy is the result of random non-disjunction which has
a low frequency of occurrence (1/250 to 1/350) though the chance increases
with age of the prospective mother. So unless the parents are older the odds
of a second Down’s child are quite remote.
On the other hand, Down’s due to a translocation has a much higher chance
of occurring a second time since there is a 25% chance for a gamete to have
the chromosome combination responsible for Down’s (chromosome 21 plus
the translocated chromosome 14) every time meiosis occurs. This would
mean there is a good chance for a second Down’s child to occur.
11. You mate two plants that you think are related because they look alike and they
have the same chromosome number. You successfully mate the plants and get
hybrid progeny but they are sterile. What type of chromosome changes could
cause the hybrid to be sterile and what would you expect to see in meiosis to
support your hypothesis?
There are several possibilities that could cause a hybrid to be sterile.
One of the parents has inversions and or translocations and the other parent
does not. This would lead to the hybrid being heterozygous for the
chromosome changes leading to partial or complete sterility. In meiosis for
inversions loops would be observed in prophase I and if the inversions were
paracentric it would be possible to see dicentric bridges and acentric
fragments during anaphase I. For translocations a cross structure of four
chromosomes would be observed in prophase I of meiosis that would open up
to either a ring of four chromosomes or a figure 8 structure of 4
chromosomes depending on how the centromeres separate at metaphase I.
A second possibility is the two species are related but their chromosomes are
homeologous so will not pair in meiosis. This would lead to an unequal
distribution of chromosomes into the gametes making them non-viable.
What would be seen in metaphase I would be only univalents and no
bivalents.
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