Answer Key Genetics Problems Chapter 10
page 235: 1-6
1
1. a.
Aa
b.
A
Aa
c.
a
A
a
AA
Aa
Aa
aa
d.
e. In c the two events must occur in a specific order. In d they can occur in either of two ways,
doubling the probability of the correct outcome.
2
AA
2. a. 1
b. 0
c.
a
aa
a
3. a. 0
A
A
Aa
Aa
Aa
Aa
3
AA
b. 1
A
AA
A
A
A
AA
AA
AA
AA
4. a. Ann = aa, Mike = Aa, Sara = Aa, Martha = aa
b.
4
Mike, Aa
c.
Ann, a
d. 1
aa
a
A
a
Aa
aa
Aa
aa
5. a. Since all six kids are normally pigmented, one would
suspect that Ralph is homozygous dominant, AA. This way
all of his offspring would receive the dominant trait for
normal pigmentation.
5.a.
Ralph, AA
Ann, a
aa
a
b. If Ralph was heterozygous for pigmentation, the probability
of he and Ann having six out of six normally pigmented
children would be 1 in 64.
5.b.
A
A
Aa
Aa
Aa
Aa
Ralph, Aa
Ann, a
aa
a
A
a
Aa
aa
Aa
aa
AaBb
6. a.
AB
Ab
aB
ab
b.
c.
d.
e.
AABb
AB
6.d.
AB
AaBb Ab
aB
ab
Ab
AB
AaBb
AB
Ab
AABB AABb
AABb AAbb
AaBB AaBb
AaBb Aabb
Ab
aB
AaBB
AaBb
aaBB
aaBb
ab
AaBb
Aabb
aaBb
aabb
p.235; 3, 5, 8-12
3. When he crossed two different true-breeding plants, he found that the offspring were all the same
as the parent with the dominant trait. The recessive trait seemed to disappear. He wondered what
happened to the recessive trait.
5. Genotypic Ratios: 1YY : 2Yy : 1yy
5
Yy
Phenotypic Ratios: 3 yellow : 1 green
Y
Yy
y
Y
y
YY
Yy
Yy
yy
8
8. a.
Aa
A
b.
, Since either parent could give A or a, the
A
chances of Aa are 2 out of 4, or ½, as seen in
the Punnett square.
c.
Aa
egg
a
sperm
a
AA
Aa
Aa
aa
, The chances of an a sperm fertilizing an A egg are 1 in 4.
d. Since either parent can give A or a, there are two ways of getting Aa. Each has a ¼ chance of
occurring. So the total probability of getting Aa is
.
9.
P1 Cross
P1
ry
rryy ry
ry
ry
RY
RrYy
RrYy
RrYy
RrYy
RRYY
RY
RrYy
RrYy
RrYy
RrYy
F1 Cross
RY
RrYy
RrYy
RrYy
RrYy
RY
RrYy
RrYy
RrYy
RrYy
F1
RY
RrYy Ry
rY
ry
F2 Ratios:
RrYy
RY
Ry
RRYY RRYy
RRYy Rryy
RrYY RrYy
RrYy Rryy
rY
RrYY
RrYy
rrYY
rrYy
Round/Yellow
Round/green
wrinkled/Yellow
wrinkled/green
ry
RrYy
Rryy
rrYy
rryy
9
3
3
1
10. Mendel’s second law is called the law of independent assortment. It says that the separation of
alleles for two different traits occur independently of each other. In the F1 cross above, the alleles
for seed color are separated independently of the alleles for seed shape. In other words, which
allele is received for seed shape is not influenced by which allele is received for seed color, or vice
versa. This allows for all possible phenotypes (four) to be seen in the offspring.
11. See diagram 10.10 on page 230
12. The purpose of a testcross is to determine the genotype of a dominant phenotype. For instance a
pea plant bearing round/yellow seeds could be RRYY, RrYY, RRYy, or RrYy. One cannot tell by
looking at the plant. A test cross with a homozygous-recessive plant would give definitive
phenotypic ratios.
ry
rryy ry
ry
ry
The genotypic ratios for the above cross are:
The phenotypic ratios are:
RY
RrYy
RrYy
RrYy
RrYy
RrYy
Ry
Rryy
Rryy
Rryy
Rryy
rY
rrYy
rrYy
rrYy
rrYy
ry
rryy
rryy
rryy
rryy
1 RrYy : 1 Rryy : 1 rrYy : 1 rryy
1 round/yellow : 1 round/green : 1 wrinkled/yellow : 1 wrinkled/green