Download doc Sample midterm 2

SAMPLE MIDTERM QUESTIONS (Prof. Schoen’s lectures)
Use the information below to answer the next two questions:
Assume that high blood pressure is inherited as an autosomal dominant trait. You
genotype the individuals in your pedigrees with many markers and successfully
localize the trait to a region on chromosome 3. A1 and A2 are two codominant
alleles of the A marker gene on chromosome 3. Our representative family and their
genotypes at the A marker are shown below. Note: Within a given generation,
individuals are numbered consecutively from left to right (e.g., I-1, I-2 for the A1/A1
and A2/A2 parents in generation I).
1. (1 point) The individual(s) in the pedigree that has (have) produced informative
meioses for mapping of the A marker and high blood pressure locus is (are):
a. I-1 and I-2
b. II-3
c. III-2, III-4, and III-5
d. III-1 and III-3
e. None of the above
Answer b. In order for an individual to produce informative meioses for mapping the
disease and marker loci, it must be doubly heterozygous and produce progeny
whose genotypes allow inference of the gamete type that was contributed by the
individual in question (e.g., whether it was a parental or recombinant type of gamete).
Given the information provided in the pedigree, we see that only II-3 fits these criteria.
Note: Genotypes of individuals in generation III provide information for mapping, but
it is not because of their meioses (i.e., we are not looking at the progeny of
generation III!).
2. (2 points) Calculate a LOD score for linkage between the high blood pressure
locus and marker A at a distance of 10 cM.
a. 0.16384
b. -0.78558
c. -2.5406
d. -0.18352
e. None of the above
Answer c. We must inspect the 5 testcross progeny present in generation III and
consider the gamete types produced by II-3. Note that parental gametes of II-3 are
either A1d or A2D (where D denotes the dominant disease allele), and recombinant
gametes are A1D or A2d. Among the 5 testcross progeny, there is one that is the
product of a parental gamete of II-3 (individual III-3), whereas the remaining 4
progeny are each the product of recombinant gametes of II-3. Thus, the appropriate
calculation of the LOD for 10 cM is:
LOD = Log [(0.054)*(0.451)/.255] = -2.5406 (where Log denotes the base 10 log of the
quantity in square brackets).
3. (1 point) In a complementation test the number of complementation groups
indicates
a.
b.
c.
d.
e.
the number of genes required for a specific phenotype.
the penetrance of the phenotype.
the number of phenotypes for a gene.
the number of chromosomes in an organism.
the quantity of gene product required for a phenotype.
Answer a.
Use the information below to answer the next two questions:
In a population, two different rare genetic diseases (extra finger and an eye disease)
are observed. In the pedigree below (taken from a family in the population), a cross
represents the occurrence of an extra finger and a black square represents the
occurrence of the eye disease:
4. (1 point) In the above pedigree, what is the most likely pattern of inheritance for
the occurrence of an extra finger?
a. autosomal dominant
b. autosomal recessive
c. X-linked dominant
d. X- linked recessive
e. Y-linked inheritance.
Answer a.
5. (1 point) In the above pedigree, what is the most likely pattern of inheritance for
the occurrence of eye disease?
a. autosomal dominant
b. autosomal recessive
c. X-linked dominant
d. X- linked recessive
e. Y-linked inheritance.
Answer d.
Use the information below to answer the next two questions:
The following three recessive markers are known in lab mice: h, hotfoot; o, obese;
and wa, waved. A trihybrid of unknown origin is testcrossed, producing the
following offspring:
hotfoot, obese, waved
357
hotfoot, obese
74
waved
66
obese
79
wildtype
343
hotfoot, waved
61
obese, waved
11
hotfoot
9
6. (2 points) The map distance between the gene pairs o and h, and h and w are
respectively:
a. 14 cM, 16 cM
b. 16 cM, 14 cM
c. 16 cM, 16 cM
d. 14 cM, 14 cM
e. None of the above
Answer c. Note that there are excesses of the triple mutant and triple wild type
progeny, suggesting linkage of the three genes and a triple heterozygote parent in
which all mutant alleles are on one of the homologs, and all wild type on the other.
Looking at the pairs of genes (o, h and h, w individually) we see that there are
79+61+11+9=160 recombinants for o and h. And there are 74+66+11+9=160
recombinants for h and w. Since there are 1000 progeny in total, this gives map
distances of 16 cM for each pair.
7. (1 point) The linear order of the genes is:
a. o,h,w
b. o,w,h
c. h,w,o
d. w,o,h
e. Both b. and c. above.
Answer a. As inferred from the low frequency of double crossovers and maps
distances above.
8. (1 point) In cocker spaniels, black color (B) is dominant over red (b), and solid
color (S) is dominant over spotted (s). The genes for these two characters are
located on different chromosomes. Suppose that a dog with genotype B/b s/s is
mated to a dog with genotype b/b S/s. What fraction of their offspring are expected
to be black and spotted?
a. 1/16
b. ¼
c. 9/16
d. 3/4
e. None of the above.
Answer b.
9. (1 point) In a cross between a female with genotype A/a B/b c/c d/d and a male
with genotype A/a b/b C/c D/d, what proportion of the progeny will be
phenotypically identical to the female parent? (Assume independent assortment of
all genes and complete dominance).
a. 1/8
b. 3/32
c. 1/64
d. 9/16
e. None of the above
Answer b.
11. (1 point) In Mendelian genetics, null mutations of different genes may interact
to produce deviations from the classic dihybrid ratio. Which of the following is most
likely due to interactions between genes in the same linear biosynthetic pathway
(i.e., PrecursorIntermediateFinal product).
a.
b.
c.
d.
e.
9:3:3:1
9:7
12:3:4
15:1
None of the above
Answer b. Interactions between null mutations of two genes in this pathway will
produce 9:7 ratio of wildtype:mutant.
Use the information below to answer the next two questions:
The recessive mutations vg (vestigial wings) and br (brown eyes) identify two
autosomal genes on the second chromosome of Drosophila melanogaster. When
females heterozygous for these genes were crossed with vestigial winged, brown
eyed males, the following classes and numbers of progeny (out of 1000) were
obtained:
Phenotype
_
wildtype wings, wildtype eyes
vestigial wings, wildtype eyes
wildtype wings, brown eyes
vestigial wings, brown eyes
Number
180
317
308
195
12. (1 point) Based upon these results, the map distance between the vg and br
genes is estimated to be:
a. 18.8 map units.
b. 31.2 map units.
c. 37.5 map units.
d. greater than 50 units because all four classes of offspring were observed.
e. None of the above.
Answer c. Note that the double heterozygote must be of the type + br / vg + , as the
excess types of progeny are not either double wild type or double mutant.
13. (2 points) Which homozygous parent stocks could be used to produce the
heterozygous females used in the cross described above?
a. wildtype wings, wildtype eyes X vestigial wings, brown eyes
b. wildtype wings, brown eyes X vestigial wings, wildtype eyes
c. wildtype wings, wildtype eyes X wildtype wings, wildtype eyes
d. vestigial wings, brown eyes X vestigial wings, brown eye
e. None of the above
Answer b. Only this pair of parental stocks could give rise to the double heterozgote
parent noted in the above answer.
Use the information below to answer the next question:
A fruit fly of genotype a+/a b+/b is crossed to another fruit fly of genotype a/a b/b.
The progeny of this cross were:
Genotype
a+/a b+/b
a/a
b/b
a+/a b/b
a/a b+/b
Number of Individuals
39
33
10
18
14. (2 point) The Chi-Square associated with the test for independent assortment of
the two genes is:
a. 20.007 (with 3 degrees of freedom)
b. 20.007 (with 1 degrees of freedom)
c. 21.360 (with 1 degrees of freedom)
d. 21.360 (with 3 degrees of freedom)
e. Cannot be calculated from the information given.
Answer b or d. As stated, the null hypothesis is independent assortment. By rejecting
the null, we might be inclined to accept the alternative, namely that the genes are
linked. But rejection of the null could also be due to differential viability. Answer “b”
takes into account the possibility of differential viability of the mutation-bearing
gametes, whereas answer “d” does not. However, I noted that there has been confusion
over this due to inconsistency in the way in which the text and MegaManual solves
these types of problems. Technically, the method that takes viability into account is
more accurate, but I will also accept the method that uses a 1:1:1:1 ratio (no viability
differences) as the null hypothesis.
15. (1 point) Non-homologous chromosomes:
a. are genetically identical.
b. are similar, but there are a few minor genetic differences between them.
c. pair with each other during meiosis I.
d. are found in mitochondria.
e. None of the above.
Answer e.
16. (1 point) An individual who is heterozygous has brittle bone disease disease (in
the mutant heterozygote an abnormal copy of the protein wraps around one or two
copies of the other and distorts the conformation of the functional trimeric
molecule). This type of mutation is called a:
a. a leaky mutation
b. a dominant-negative mutation
c. recessive mutation
d. haplo-insufficient mutation
e. haplo-sufficient mutation
Answer b. Note: “Haplo-insufficient mutation” is an incorrect answer. While this is a
dominant mutation, the phenotypic basis does NOT arise because two functional doses
of an enzyme are required for completion of a biosynthetic pathway. Rather, the
mechanism is that of “poisoning” of one protein unit by another (a dominant
negative), as discussed in the text and class (and contrasted with haplo-insufficient
mutations) under the topic of dominant mutations (see Lecture 8).
17. (1 point) A man with blood type "B" marries a woman with blood type "B".
Which of the following blood types might you expect to see in their children?
a. Type A and B would be possible.
b. Only type O would be possible.
c. Only type B would be possible.
d. Types B and O would be possible.
e. Only Type AB would be possible
Answer d.