Download C. Would you expect cells treated with methotrexate to produce

Life Sciences 1b Problem Set 3 ANSWER KEY
Due in TF dropbox (Science Center 2nd floor), Monday Feb 22 at 12:00pm.
1.
(5 points) The African pygmy hedgehog, Atelerix albiventris, has a diploid chromosome number of
90, including one pair of sex chromosomes.
A.
How many chromosomes would you expect to find in a pygmy hedgehog somatic cell during
metaphase of mitosis? How many chromatids? Would the chromatids be identical? Explain your
reasoning.
+1.5: 90 chromosomes (+0.5), each containing 2 chromatids, for a total of 180 chromatids (+0.5).
The chromatids would be identical (+0.5) (except for any mutations that occurred during DNA
replication), since no crossing over occurs during mitosis.
B.
How many chromosomes would you expect to find in a pygmy hedgehog somatic cell during
anaphase of mitosis? How many chromatids? Explain your reasoning.
+1: 90 x 2 = 180 total chromosomes (+0.5), 0 chromatids because they have separated (+0.5)
C.
How many chromosomes would you expect to find in a pygmy hedgehog germ cell during
metaphase of meiosis I? How many chromatids? Would the chromatids be identical? Explain
your reasoning.
+1.5: 90 chromosomes (+0.5), each containing 2 chromatids, for a total of 180 chromatids (+0.5).
The chromatids would not be identical, because crossing over occurs during prophase I of
meiosis (+0.5)).
D.
How many chromosomes would you expect to find in a pygmy hedgehog germ cell during
metaphase of meiosis II? Explain your reasoning.
+1: 45 chromosomes (+0.5) the homologous pairs have separated, and each cell got one of each
homolog (+0.5). Each chromosome has two sister chromatids.
2.
(5 points) Methotrexate is a drug that can inhibit cell division by preventing the synthesis of purine
bases. Cells treated with methotrexate arrest at a particular stage of the cell cycle.
A.
During which stage of the cell cycle would you expect cells treated with methotrexate to arrest?
Briefly explain.
+1;Without purine bases, nucleotides cannot be made and DNA synthesis could not occur (+0.5).
Cells may initiate S phase, but would not be able to complete it because they cannot make DNA
without being able to make new bases, so they would arrest before or during S phase (If cells
recognize the lack of purines before initiating DNA synthesis, they could arrest at the G1/S
checkpoint. It is also possible that cells would initiate S phase and get “stuck” partway through
DNA replication) +0.5 for explanation.
B.
Below are histograms (graphs) showing the numbers of cells containing different amounts of DNA
in three different samples: cells treated with two different drugs (A and B), and cells that have not
been treated with a drug. The untreated cells are progressing through the cell cycle at a normal
rate, while the cells that have been treated with drugs tend to arrest at different stages in the cell
cycle.
Untreated cells
Untreated cells
x
2x 2x
x
content
DNADNA
content
# of cells
# of cells
# of cells
S
# of cells
# of cells
Treated with Drug B
Treated with Drug B
G2
G1
# of cells
Treated with Drug A
Treated with Drug A
x
x 2x
2x
DNA content
DNA content
x
x2x
2x
DNA content
DNA content
On the histogram of “untreated cells” indicate which bars in the graph represent cells in each of
the following stages of the cell cycle: G1, S, and G2. If there are bars that you cannot assign to
one of these stages, do not label them.
+1.5;+0.5 per stage correctly labeled.
C.
Would you expect cells treated with methotrexate to produce results more similar to those shown
for Drug A or Drug B? Briefly explain your reasoning, including how the data in the graph is
consistent with the effect of methotrexate on cell cycle progression.
+1;Drug A (+0.5) – the cells cannot complete DNA synthesis because they are unable to make
purine-containing nucleotides to use during S phase, thus there are many cells in G1 and early S
(with a small amount of DNA replicated), but not many in later S or G2/M, with 2x DNA content
+0.5.
D.
Which of the following drugs could be drug B? Briefly explain your reasoning for each.

Vinblastine, which binds to tubulin and inhibits microtubule assembly
+1.5;
Yes, as in this treatment more cells have 2x DNA, showing that the process that was blocked is
post-DNA synthesis. Vinblastine would inhibit mitosis, during which cells have 2x DNA (+0.5).

PPMP, a cytokinesis inhibitor
Yes, as in this treatment more cells have 2x DNA, showing that the process that was blocked is
post-DNA synthesis. PPMP would inhibit cytokinesis, which takes cells back to 1x DNA from 2x.
(+0.5)
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3.
(8 points) A particular species of scorpion has two pairs of homologous chromosomes, and shown
below are the karyotypes of a normal male and a normal female scorpion as well as selected
offspring from a mating between them. The larger chromosome is chromosome I, and the smaller
chromosome is chromosome II. Remember that karyotypes are made using metaphase
chromosomes, so each chromosome shown contains two sister chromatids, though they cannot
be distinguished from one another, as they are held together tightly.
Karyotype of
female parent:
Karyotype of
male parent:
Karyotypes of selected offspring:
1
2
3
4
5
A.
For this species of scorpions, n = 2 (+0.5) and 2n = 4 (+0.5)
B.
For each of the offspring (1-5), identify whether the karyotype is normal or if there is an
abnormality. If there is an abnormality, identify the abnormality (e.g., monosomy chr. II), and in
which parent and which phase of meiosis (I or II, or either) the error could have occurred. +5.5;
Offspring
C.
Normal (yes/no)?
Abnormality (if
applicable)
Parent and phase of meiosis (I or II)
in which abnormality occurred (if
applicable)
1
Yes (+0.5)
2
No (+0.25)
Trisomy I (+0.5)
Meiosis I in female (+0.5)
3
No (+0.25)
Triploid (+0.5)
Meiosis I in female (+0.5)
4
No (+0.25)
Trisomy II (+0.5)
Meiosis II in male (+0.5)
5
No (+0.25)
Monosomy I (+0.5)
Meiosis I or II in male (+0.5)
Generally, offspring with chromosomal abnormalities such as these are not viable – they die
early in development. Individuals with some chromosomal abnormalities, however, are able to
survive. What factor determines the extent of the defect caused by a particular chromosomal
abnormality (i.e., why are humans with trisomy 21 sometimes able to survive, while individuals
with trisomy 1 are not)? Why is this factor so important?
+1.5; The extent of the defect caused by a particular chromosomal abnormality is dependent on
the number of genes/size of chromosome, as larger chromosomes tend to have more genes
(+0.5). This will result in the production of too much (trisomy) or too little (monosomy) gene
products from that chromosome, which has a greater impact when more genes are affected.
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The number of genes is important because of gene dosage effects – the more genes affected,
the more likely that changes in the dosage (relative proportions of gene products) of one or more
of them will lead to catastrophic defects in development, and inviable offspring. For example, if
you have an extra copy of the chromosome containing a gene for a transcriptional activator, you
will make even more of that activator relative to other transcriptional regulators, disrupting the
regulation of many different gene (+1 for an explanation)
4.
(8 points) In plants, the NBS-LRR genes mediate disease resistance. Many of these genes have
highly similar sequences and are located adjacent to one another in the genome.
Consider the following cluster of NBS-LRR genes from a specific locus in Arabidopsis thaliana.
There are six NBS-LRR genes at this site in the genome (NBS-LRRA – NBS-LRRF), and the diagram
below shows their relative positions on the homologous chromosomes of an individual who is
heterozygous for all six genes (NBS-LRRA/NBS-LRRa, NBS-LRRB/NBS-LRRb, etc.).
NBS-LRRA
NBS-LRRB
NBS-LRRC
NBS-LRRD
NBS-LRRE
NBS-LRRF
NBS-LRRa
NBS-LRRb
NBS-LRRc
NBS-LRRd
NBS-LRRe
NBS-LRRf
Below are the genotypes of three gametes produced by this individual. Some of the gametes
have missing or extra NBS-LRR genes due to unequal crossing over during meiosis. For each
gamete shown (I) determine between which genes the crossover occurred on each
chromosome and (II) write out the genotypes of the other three products of meiosis.
NBS-LRRa
A.
NBS-LRRb
NBS-LRRc
NBS-LRRd
NBS-LRRC
NBS-LRRD
NBS-LRRE
NBS-LRRF
+0.5 total for correctly identifying that there will be two non-recombinant gametes produced from
each crossover event.
Crossing over occurred between NBSLRRd/NBSLRRe and NBSLRRB/NBSLRRC. (+1 for crossover site)
The other gametes would contain (1) ABCDEF (2) abcdef and (3) ABef (+1 for product of unequal
crossing over)
NBS-LRRa
B.
NBS-LRRb
NBS-LRRc
NBS-LRRd
NBS-LRRF
Crossing over occurred between NBSLLRRd/NBSLRRe and NBSLRRE/NBSLRRF (+1 for crossover site)
The other gametes would contain (1) ABCDEF (2) abcdef and (3) ABCDEef (+1 for product of
unequal crossing over)
C.
NBS-LRRA
NBS-LRRB
NBS-LRRc
NBS-LRRd
NBS-LRRe
NBS-LRRf
Crossing over occurred between NBSLRRB/NBLSRRC and NBSLRRb/NBSLRRc (+1 for crossover site)
The other gametes would contain (1) ABCDEF (2) abcder and (3) abCDEF (+1 for product of
crossing over).
D.
Which of the gametes from A, B, and C would you predict would produce a next-generation
individual with reduced resistance to disease as a result of the loss of one or more NBS-LRR genes?
A: the gamete with only ABef
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B: the gamete with only abcdF
C: none of them
+1.5; 0.5 per answer.
5.
(9 points) You are interested in behavioral genetics, and have selected zebrafish as your model
system. You are studying how quickly fish respond to a flash of light, and have identified several
mutants that respond more slowly than wild-type fish.
You take three females who exhibit this delayed-response phenotype, and cross each with a
homozygous wild-type male. In the progeny, you observe the following numbers of normal and
slow-responding offspring:
# of normal offspring
# of offspring with delayed
response phenotype
Cross 1
21
32
Cross 2
42
0
Cross 3
0
37
A.
B.
For each cross, identify whether the mutant phenotype is dominant or recessive, and the
genotype of the female from the cross. Define any symbols that you use.
+4.5;Let the gene that contributes to the “delayed response” phenotype = D; dominant alleles are
indicated in capital letters, recessive alleles in lowercase.
Cross 1: mutant phenotype is dominant (+0.5), if D = mutant allele and d = WT (wild-type), female
was Dd (+1)
Cross 2: mutant phenotype is recessive (+0.5), if d = mutant allele and D = WT, female was dd (+1)
Cross 3: mutant phenotype is dominant (+0.5), if D = mutant allele and d = WT, female was DD (+1)
Do you think that the mutation in the female from cross 1 is more likely to be a loss of function
mutation or a gain of function mutation? What about cross 2? Briefly explain how you came to
each decision in the space provided.
+2; Generally speaking, gain of function mutations tend to be dominant. Loss of function mutations
are dominant in the case of haploinsufficiency, otherwise they tend to be recessive. So the
mutation in cross 1 could either be a be a gain of function or a haploinsufficient LOF while the
mutation in cross 2 would more likely be a loss of function.
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C.
You decide to study one of these mutations further by making a transgene that contains the
mutation using the strategy discussed in class (inject a plasmid containing the transgene into 1celled embryos). For this purpose, you select a dominant mutation. Why is it necessary for the
mutation to be dominant in order for this strategy to work?
+1.5; This transgenic strategy relies on adding extra copies of the gene, so you need to use a
dominant allele if you want to see the phenotype since the wild-type alleles will still be present
and would mask expression of a recessive phenotype.
D.
If you cross two of the offspring from cross 3 with each other, what proportion of individuals in the
next generation would you expect to display the delayed-response phenotype? Show your work
or explain your reasoning.
+1: Dd x Dd = ¼ DD, ½ Dd, and ¼ dd – both DD and Dd will show the phenotype, so ¾
6.
(5 points) Cystic fibrosis is inherited in an autosomal recessive manner. An unaffected male has
two children with an unaffected female. One of their children, Chen is unaffected by the disease;
the other child, Chao, has cystic fibrosis
A.
What is the genotype of Chen and Chao’s mother? Briefly explain your answer.
+1: Cc (+0.5) – She is unaffected therefore cannot be homozygous recessive cc and since she has
affected children she must be a carrier (+0.5).
B.
What is the probability that Chen is a carrier for cystic fibrosis? Show your work or explain your
reasoning?
+1: 2/3 (+0.5 for probability): Chen is unaffected and therefore their genotype is definitely not cc.
There are three other possible genotypes: Cc, cC or CC. 2/3 of which are carrier genotypes (+0.5
for explanation/working).
C.
Chen has a child with an unaffected female named Lin. Lin has an affected brother. What is the
probability that Chen and Lin’s child will be an unaffected girl? Show your work or explain your
reasoning.
+3:
p both parents are carriers: 2/3 (Cc) x 2/3 (Cc) = 4/9
p Chen is a carrier & Lin is not: 1/3 (CC) x 2/3 (Cc) = 2/9 +
p Lin is a carrier & Chen is not: 1/3 (CC) x 2/3 (Cc) = 2/9 +
p neither parents are carriers: 1/3 (CC) x 1/3 (CC) = 1/9
=5/9
Probability of having an unaffected child = (4/9 x 3/4) + (5/9 x 1) = 1/3 + 5/9 = 8/9 (+1.5 for
correctly determining probability of unaffected child)
Probability of having an unaffected girl = 8/9 x 1/2 = 4/9 (+0.5 for multiplying by ½ for a girl)
+1 for correct application of product and sum rule.
Alternative solution:
1-(p of affected child: 2/3 * 2/3 * 1/4 = 1/9) = 8/9 * 1/2 (probability of a girl) = 4/9
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