Download Problems 10

BS 50—Genetics and Genomics
Week of Oct 10
Additional Practice Problems for Section
Answer Key
1. Two purebreeding strains of mice are crossed to produce a mouse heterozygous for dominant
and recessive alleles of three linked genes (A vs. a, B vs. b, E vs. e). A series of F1 triply
heterozygous mice are testcrossed and the resulting progeny were:
ABE
8
ABe
342
AbE
100
Abe
62
aBE
49
aBe
94
abE
317
abe
14
986 total progeny
Justify your answer to each of the following questions:
a. What are the genotypes of the purebred parents of an F1 triple heterozygote?
8 points total. Genotypes ABe and abE are the most frequent classes, which
indicates the parents. 4 points deducted if no justification given.
b. What is the map order of the three genes? (map distances are not required)
12 points total. The rarest classes indicate the double crossover, and thus the
map order A-E-B. 6 points deducted if no justification given.
2. Genes A, B, and R are linked to one another. A genetic map of the locations of these genes is
shown below:
A
R
B
|----------30cM-----------|----10cM---|
cM = m.u. = 1% recombination
each part was worth 10 points. Being quantitative was essential. For either part a or b,
deducted half if the student was on the right track but left off a factor (e.g., the 1/2).
a. You cross a pure-breeding A R strain with a pure-breeding a r strain, and then testcross the F1
progeny. What percentage of the testcross progeny do you expect to be phenotypically A r?
a) A r = 1/2 of recombinant progeny = 1/2 (30%) = 15%
b. You cross a pure-breeding A R B strain with a pure-breeding a r b strain and then testcross
the F1 progeny. What percentage of the testcross progeny do you expect to be phenotypically
A R B?
b) The probability of obtaining a parental chromosome (non-recombinant) in this region is
the product of the probability of no recombination between A & R (70%) and the
probability of no recombination between R & B (90%). This would mean that (70%)(90%)
= 63% should be “parental” and 1/2 (63%) = 31.5% would be expected for a single parental
class.
Question 3 (18 pts)
8 pts A) The test cross becomes:
A/a ; B/b ; C/c x a/a ; b/b ; c/c
so the probability of obtaining an offspring with the same phenotype as the Fl parents is:
1/ x 1/ x 1/ = 1/
2
2
2
8
10 pts B) The test cross is: A/a ; B/b ; C/c ; De/dE x a/a ; b/b ; c/c ; de/de. The D and E genes
are linked. To obtain a progeny of this cross with the same phenotype as the F1 parent requires
that the recombinant chromosome DE is inherited. Since there are 20 cM between these two
genes, this probability is 10% . So the probability of obtaining an offspring with the same
phenotype as the Fl parents is:
1/ x 1/ x 1/ x 10% = 1/
2
2
2
80
Problem 4
4 pts a) No. Independent assortment of three recessive markers would be expected to give a
27:9:9:9:3:3:3:1 ratio (3:1 × 3:1 × 3:1) in the F2....or, there are many more yellow-bodied and
white-eyed (or black-bodied and red-eyed) than yellow-bodied and red-eyed or white-eyed and
black-bodied. (It is clear that these two markers are linked by observing their inheritance.)
8 pts b) Let H = normal-winged; h = hairy-winged; R = red-eyed; r = white-eyed; B = blackbodied; and b = yellow-bodied.
The data indicate that r and b (and R and B) are linked and further that these loci are on the X
chromosome. The hairy locus is unlinked to these genes and also inherited as an autosomal trait.
So the parents are:
h/h; RB/RB females and H/H; rb/Y males
6 pts c) The F1 flies are H/h; RB/rb females and H/h; RB/Y males
Problem 5
a) W; it remains outside the cell
b) X = RNA; Y = single-stranded DNA; Z = single- or double-stranded DNA
c) Z; only one that could be double-stranded DNA
6)
L/L
H/L H/H
generation 0: 100% 0%
0%
generation 1: 0%
100% 0%
generation 2: 50% 50% 0%
Problem 6
The following represent various stages of cell division in cells from the same individual
organism.
A)
B)
D)
C)
E)
What stage of cell division is represented by each drawing? Choose from the following list:
a) meiotic metaphase II
b) mitotic anaphase
c) mitotic metaphase
d) meiotic telophase II
e) meiotic anaphase I
f) 3
10 pts.
a-e (8 pts):
f) (2 pts)
0 wrong = 8
1 wrong = 7
2 wrong = 5
3 wrong = 3
4 wrong = 2
5 wrong = 0