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BIOLOGY 202
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Dr. Schoen’s Questions (64 points)
1. (3 points) The following is a series of statements about homologous chromosomes:
i) They always contain the same genes.
ii) They always contain the same alleles.
iii) They occur together in diploid cells.
iv) They occur together in haploid cells.
For statements i through iv, which are true statements and which are false?
a.) i. True, ii. False, iii. True, iv. False
b.) i. False, ii. True, iii. True, iv. False
c.) i. True, ii. False, iii. False, iv. True
d.) i. False, ii. True, iii. False, iv. True
e.) None of the above
Answer: a. I will also accept “e”, as if we define X and Y sex chromosomes as
being “homologs” (which some genetcists do), then none of the answers are correct.
2. (6 points) Suppose that coat color in mice is governed by the “B” locus. Black coat
color (BB or Bb) is dominant to brown (bb). A black mouse was crossed with a brown
mouse and all three progeny were black. How can you explain these results?
a.) The black parent must have had the genotype BB.
b.) The black parent must have had the genotype Bb, but all three of its progeny
were from fertilizations with B gametes.
c.) There is not enough information to be certain of the genotype of the black
parent.
d.) Another locus must be involved in coat color determination.
e.) None of the above.
Answer c. Note: The probability that all three progeny from a Bb x bb cross are
black, is equal to 1/8 (this is sufficiently large such that it could occur by chance).
Thus even though all progeny are black, we can not with reasonably high probability
be condifent that Bb or BB are genotype of the black parent (Remember that we
exclude null hypotheses when the probability of acceptance is less than 1/20). In other
words, the probability that a Bb black parent produces all brown progeny when crossed
with a brown parent is considered reasonably high (one out of eight, which is well
within the margin of chance alone being responsible for the outcome when the black
parent is Bb.) So it is the result (the data themselves and not the question) that is
ambiguous in terms of resolving the parental genotype, leaving only answer “c” as
correct response. This was intended to be a challenging question.
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3. (3 points) How many different types of haploid gametes can an individual with the
genotype AaBBccDdEeFf form (assume independent assortment)?
a) 4
b) 12
c) 16
d) 64
e) 256
Answer c. We have four segregating and independently assorting genes, hence
the correct calculation is 2 raised to the 4th power = 16 genotypes.
4. (3 points) What is the mechanism that ensures Mendel’s Law of Equal Segregation?
a)
b)
c)
d)
e)
Segregation of sister chromatids during meiosis II
The formation of the kinetochore
Pairing of the dyads into tetrads
Formation of the chiasmata
Segregation of homologous chromosomes during meiosis I
Answer e. Segregation of alternative alleles occurs in meiosis I, when the
paired sister chromatids are passed on to separate daughter cells (that is AA go to one
daughter cell, while aa go to the other). During meiosis II, these daughter cells
produced in meiosis I divide (which is not segregation).
5. (3 points) A rare dominant trait, when exhibited in men, is transmitted to half their
sons and to half their daughters. The gene for this trait is carried:
a)
b)
c)
d)
e)
on the X chromosome.
on an autosome.
on the Y chromosome.
in the mitochondria.
None of the above.
Answer: b.
6. (6 points) The following pedigree shows the expression of a dominant mutation. What
is the probability that the child of IIc and IId will express this trait?
I
II
a
b
c
d
?
e
f
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a) ¼
b) ½
c) 2/3
d) ¾
e) 1
Answer: d (Note: IIc and IId must be heterozygous.)
7. (6 points) Huntington’s disease is a late-onset disease caused by a single,
dominant, autosomal mutation. The following pedigree is for a family with a
history of Huntington disease. Those individuals who are already suffering
from the disease are shaded black. However, some additional individuals in
generations II and III also have the Huntington allele and will develop
Huntington disease but have not yet shown symptoms. Assume that
individuals marrying into the family have no history of Huntington disease
(that is, they are homozygous recessive for the gene). Also assume that the
diseased male in generation I is heterozygous for the disease gene. If
individuals IIIg and IIa had a child together, what is the probability that the
child would develop Huntington disease?
I
II
a
b
c
e
d
f
g
k
l
III
a
b
c
d
e
f
g
h
i
j
m
a) ⅛
b) ¼
c) ½
d) 1
e) 0
Answer: a. Since this is a genetic disease, we must treat it as rare in the
population. This translates into the assumption that individual Ia (the
father at the top of the pedigree) is heterozygous, while individual IIa (an
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individual that “joins the extended family from outside the pedigree) is
homozygous wild type. Because the disease is late onset (and thus not
necessarily expressed in all individuals at the time the pedigree was
constructed), we have to assume that IIc could be either a carrier (with
probability ½) or a non-carrier (also with probability ½). The correct
approach when calculating the probability that IIIg is also a carrier is to
take the probability that IIc is a carrier, and multiply that probability by the
probability that IIc passed on the deleterious allele to IIg, or in other words,
the probability that IIIg is a carrier = ½ x ½ = ¼. Finally, to calculate the
probability that a Huntington offspring would occur when individuals IIIg
and IIa have a child together, we have to multiply the probability that IIIg is
a carrier by ½ for an overall probability of 1/8.
8. (3 points) The pedigree below shows the inheritance pattern of a rare trait. Which of
the following modes of inheritance best describes the mode of inheritance of the trait
presented in this pedigree?
a) Autosomal dominant
b) Autosomal recessive
c) Cytoplasmic Inheritance
d) X-linked recessive
e) Y-linked
Answer: d.
9. (6 points) The pedigree below shows the inheritance pattern of a common trait. Which
of the following modes of inheritance best describes the mode of inheritance of the trait
presented in this pedigree?
a) Autosomal dominant
b) Autosomal recessive
c) Cytoplasmic Inheritance
d) X-linked recessive
e) Y-linked
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I
II
III
a
IV
b
c
d
e
NOTE: No information available for generation IV
Answer: c. While the pedigree could be consistent with autosomal dominance,
there is very strong evidence that suggests otherwise .... in particular, the fact that in
the second generation we have a father with the trait, and a mother without it, giving
rise to five offspring who also lack the trait. This could happen (with autosomal
dominance) only if the father was heterozygote, but the chances of it happening would
then be 1/2 raised to the 5 ( = 1/32, which is considered unlikely). A similar
probabilistic argument can be raised against genotype assignments involving an
autosomal recessive basis. When faced with the competing possibilities that could
explain the data and asked to chose the best answer, the choice of cytoplasmic
inheritance is easier to accept, as all progeny of affected mothers are themselves
affected, while all progeny of unaffected males are unaffected.
10. (3 points) A trihybrid cross is a cross between two individuals who are heterozygous
for three genes. For example: AaBbCc x AaBbCc. Assuming these three genes are
unlinked and assorting independently (and each gene affects a differnent trait), what
segregation ratio would be observed?
a) 9:3:3:1
b) 1:1:1:1
c) 16:9:9:3:3:1
d) 27:9:9:9:3:3:3:1
e) 12:9:3:1
Answer: d
11. (3 points) Suppose you crossed true-breeding Drosophila females of wild type red
eye appearance and normal wing length with true-breeding males showing two
autosomal recessive traits: purple eyes and vestigial wings. If 1000 progeny of the
testcross are distributed in the following phenotypic classes (see below), what is the
estimated genetic distance between the loci for eye color and wing length? (Note: the
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heterozygote parent in the testcross was female--recombination does not occur in
Drosophile males).
Wild type
Purple eyes, vestigial wings
Red eyes, vestigial wings
Purple eyes, normal wings
a)
b)
c)
d)
e)
441
484
30
45
75 cM
0.75 cM
7.5 cM
750 cM
cannot be determined from the data
Answer: c Note: A testcross, by definition, involves a cross between a heterozygous
and a homozygous recessive (i.e., tester) parent.
12. (6 points) Genes a, b, and c are known to be in that linear order in an organism. A
testcross is done that yields 20 double crossover progeny out of a total of 1000 offspring.
Assume that there is no interference. Which map(s) below are consistent with the data?
(Note: “mu” stands for “map units”).
Map 1 a
20 mu
b
Map 2 a
10 mu b
20 mu
Map 3 a
14.14 mu
b
10 mu c
c
14.14 mu
c
a.) Map 1 only
b.) Map 2 only
c.) Maps 1 and 2
d.) All three maps
e.) None of the three maps
Answer: d (I will also accept “c” for people who did not round off their calulations in
the case of Map 3).
13. (6 points) Females homozygous for the recessive mutant characters a, b, and c were
crossed with wild-type males from a pure-breeding stock. A testcross was then done by
taking the F1 females (all wild-type in appearance) and crossing them with triple mutant
males (a, b, c) taken from a pure-breeding stock. The following offspring were counted
(Note: lower case letter indicates a mutant phenotyoe, while “+” indictes wild type):
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Phenotype
a, b, c
a++
b++
c++
a, b +
a, c +
b, c +
wild-type
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Number
280
77
23
125
127
25
75
268
1000
Here are some possible recombination maps.
Map 1:
a_______b___________c
20 cM
30 cM
Map 2:
c_______b___________a
20 cM
30 cM
Map 3:
c___________b_______a
30 cM
20 cM
Map 4:
a___________b_______c
30 cM
20 cM
Which answer below is correct?
a.) Map 1 is consistent with the data presented, and there is no evidence of interference.
b.) Maps 1 and 3 are consistent with the data presented, and there is evidence of
interference.
c.) Maps 2 and 4 are consistent with the data presented, and there is evidence of
interference.
d.) Map 1 is consistent with the data presented, and there is evidence of interference.
e.) Map 2 is consistent with the data presented, and there is no evidence of interference.
Answer: b
14. (3 points) An individual who is heterozygous for the alleles that cause Kjer’s optic
atrophy (a disease that leads to reduced visual acuity) fully expresses the disease. This
type of mutation is called a:
a) null mutation
b) leaky mutation
c) recessive mutation
d) haplo-insufficient mutation
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e) haplo-sufficient mutation
Answer: d (I will accept “e”, as the wording above could be clearer. Technically
speaking, we should refer to the wild type allele as “haplo-insufficient” in this case.)
15. (4 points) Suppose that in your lab you occasionally get mice that have some sort of
bizarre mutation—bent ears. You determine that the bent-ear trait is recessive. You then
find out that someone else in your lab has isolated mice with bent ears as well, and that
they have also determined that bent ears is recessive. Imagine that you cross one of your
bent-eared mice with one of your lab mate’s. You then intercross these offspring, forming
an F1 generation. In the F1 generation a ration of 9:7 (normal to bent) is observed. What
would explain these findings?
a.)
Gene 1 codes for product
Gene 2 codes for product
involved step below:
involved in step below:
Compound A
→
Compound B
→ normal ear shape
b.) Complementation.
c.) Independence of gene action.
d.) Suppression.
e.) None of the above
Answer: This question was accidentally combined with another from a previous year,
and actually has two answers. Answers a or b are acceptable, but I will also accept
“e”. Note: There was also a misprint in the 2nd to last sentence, which should have
read “In the F2 generation a ratio of 9:7 (normal to bent) is observed” instead of “In
the F1 generation a ratio of 9:7 (normal to bent) is observed”. Most students caught
this and answered accordingly. If, however, you read the question as is, the only other
acceptable response is “e”. Independence of gene action applies to separate traits, not
a single one, and so is not relevant. Suppression is not compatible with these results.