Download Genetics Practice Problems Key

Genetics Practice Problems Key
1. How many different types of gametes can be formed by individuals with the following
genotypes? What are the genotypes in each case?
a) AaBBcc 2 ABc, aBc
b) AaBbCc 8 ABC, AbC, Abc, aBC, aBc, abC, ABc, abc
2. A male donkey with a diploid chromosome number of 62 is mated with a zebra female with a
diploid chromosome number of 44. This mating produced a hybrid offspring called a zedonk.
a) Predict how many chromosomes you would observe in the somatic cells of the zedonk.
Make sure to explain or show how you determined your predicted number.
Correctly determined that zedonk would have 53 chromosomes 31 from Mom and 22
from Dad. Student understood that the parental gametes would have a haploid number of
chromosomes and would donate their specific chromosome number to their offspring,
so 31+22 = 53 in somatic cell.
b) Do you predict that this zedonk would be able to from viable gametes? Using your
understanding of meiosis, explain why or why not.
Student understood that errors would occur in meiosis I when homologs separate (1 pt).
The zedonk would have several chromosomes that would not have a homolog and thus
would not separate correctly during meiosis (1 pt). So the gametes of the zedonk would
not have a representative of every necessary chromosome, which would make them
inviable (1 pt).
3. A male cat has the short hair, a stubby tail, and extra toes. A female cat has long hair, a long
tail, and extra toes. The genes and alleles are as follows:
L = short hair, l= long hair
M= stubby tail, m = long tail
P = extra toes, p = normal number of toes
The two cats have kittens. The first kitten has long hair, a long tail, and no extra toes. The
second kitten has short hair, a stubby tail, and extra toes. The third kitten has short hair, a long
tail, and no extra toes.
What is the genotype of the father? LlMmPd
4. The ability to taste the chemical phenylthiocarbamide is an autosomal dominant phenotype,
and the inability to taste it is recessive. If a taster woman marries a taster man who in a
previous marriage had a nontaster daughter, what is the probability that their first child will be:
a. A nontaster girl
If mom is homozygous for Taster allele: ½ chance of having a girl * 0/0 chance of having
nontaster child= zero chance of having a nontaster giri.
If mom is heterozygous for Taster allele: ½ chance of having a girl * 1/4 chance of having
taster child= 1/8 chance of having a non-taster giri.
b. A taster girl
If mom is homozygous for Taster allele: ½ chance of having a girl * 1/1 chance of having
taster child= 1/2 chance of having a taster giri.
If mom is heterozygous for Taster allele: ½ chance of having a girl * 3/4 chance of having
taster child= 3/8 chance of having a taster girl.
c. A taster boy
If mom is homozygous for Taster allele: ½ chance of having a boy * 1/1 chance of having
taster child= 1/2 chance of having a taster boy.
If mom is heterozygous for Taster allele: ½ chance of having a boy * 3/4 chance of having
taster child= 3/8 chance of having a taster boy.
5. Four genes (A, B, C, and D) are on the same chromosome. The recombination frequencies
are as follows: A-B: 19%; B-C: 14%; A-C: 5%; B-D: 2%; A-D: 21%; C-D: 16%. Based on this
information, what is the correct sequence of genes? Please show all distances among all genes.
Correct map of chromosome
Labeled all map distances correctly.
21 A—5—C-------14-------B-2-D
100
19 80
100
16 80
85
90
85
90
75
60100
85
75
72.5
60
100
85
95
72.5
77.5
80
85
90
75
60
6. The recombination frequencies for 6 genes on a single chromosome of the silkworm Bombyx
mori are shown in the table below. Construct a genetic map that includes all of these genes.
Gr
Rc
S
Y
P
oa
Gr
25
1
19
7
20
Rc
25
26
6
32
5
S
1
26
20
6
21
Y
19
6
20
26
1
P
7
32
6
26
27
oa
20
5
21
1
27
-
Start by figuring out:
Gr is between S and Rc
Y is between Gr and Rc
Continue. At each of the subsequent steps there are two alternatives, one of which can
be rejected on the basis of the map distances.
The final map is P-S-Gr-Y-oa-Rc
7. The offspring of one mated pair of mammals included three males, all of which showed an Xlinked recessive trait. The fourth offspring was a female. Consider whether the following
statements about her are true or false. What is your reasoning?
Even if the father shows the trait, she will not. She could if she inherited the recessive trait
from Mom as well
She will definitely show the trait. Not if she inherits the dominant allele from mom
If the father does not show the trait, she will not. True, she could be a carrier though
8. Albinism is inherited as an autosomal recessive. For the figure below, assume that all
individuals not descended from the first couple are homozygous dominant (normal). On the
figure below, circle all the carriers of albinism.
Answer:
Other good problems at the end of chapter 13 page 255: Genetic Problems 1,2,3,5,6, 7,
8,11,14