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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.
3. (3 points) How many different types of haploid gametes can an individual with the genotype
AaBBccDdEeFf form (assume independent assortment)?
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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 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
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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
I
II
III
a
b
c
d
e
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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 heterozygote parent in the testcross was female-recombination does not occur in Drosophile males).
Wild type
Purple eyes, vestigial wings
Red eyes, vestigial wings
441
484
30
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Purple eyes, normal wings
a)
b)
c)
d)
e)
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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):
Phenotype
a, b, c
a++
b++
c++
a, b +
Number
280
77
23
125
127
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a, c +
b, c +
wild-type
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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
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
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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.