Download 2. Assuming homozygosity for the normal gene, the mating is A/A · b

2.
Assuming homozygosity for the normal gene, the mating is A/A · b/b × a/a · B/B.
The children would be normal, A/a · B/b (see Problem 12).
4.
Growth will be supported by a particular compound if it is later in the pathway
than the enzymatic step blocked in the mutant. Restated, the more mutants a
compound supports, the later in the pathway it must be. In this example,
compound G supports growth of all mutants and can be considered the end
product of the pathway. Alternatively, compound E does not support the growth of
any mutant and can be considered the starting substrate for the pathway. The data
indicate the following:
a. and b.
E
A
C
B
D
G
5
4
2
1
3
vertical lines indicate the step where each mutant is blocked
c.
A heterokaryon of double mutants 1, 3 and 2, 4 would grow as the first would
supply functional 2 and 4, and the second would supply functional 1 and 3.
A heterokaryon of the double mutants 1, 3 and 3, 4 would not grow as both
are mutant for 3.
A heterokaryon of the double mutants 1, 2 and 2, 4 and 1, 4 would grow as
the first would supply functional 4, the second would supply functional 1,
and the last would supply functional 2.
5.
6.
a.
If enzyme A was defective or missing (m2/m2), red pigment would still be
made and the petals would be red.
b.
Purple, because it has a wild-type allele for each gene, and you are told that
the mutations are recessive.
c.
9
3
3
1
d.
The mutant alleles do not produce functional enzyme. However, enough
functional enzyme must be produced by the single wild-type allele of each
gene to synthesize normal levels of pigment.
a.
If enzyme B is missing, a white intermediate will accumulate and the petals
will be white.
b.
If enzyme D is missing, a blue intermediate will accumulate and the petals
will be blue.
c.
M1/– ; M2/–
m1/m1 ; M2/–
M1/– ; m2/m2
m1/m1 ; m2/m2
P
purple
blue
red
white
b/b ; D/D × B/B ; d/d
108 Chapter Six
d.
F1
B/b ; D/d
purple
P
b/b ; D/D × B/B ; d/d
F1
B/b ; D/d × B/b ; D/d
F2
9 B/– ; D/–
3 b/b ; D/–
3 B/– ; d/d
1b/b ; d/d
purple
white
blue
white
The ratio of purple : blue : white would be 9:3:4.
7.
The woman must be A/O, so the mating is A/O × A/B. Their children will be
Genotype
1/4 A/A
1/4 A/B
1/4
1/4
9.
a.
A/O
B/O
Phenotype A
AB
A
B
The original cross and results were
P
long, white × round, red
F1
oval, purple
F2
9 long, red
19 oval, red
8 round, white
15 long, purple32 oval, purple 16 round, purple
8 long, white
16 oval, white
9 round, red
32 long
67 oval
33 round
The data show that, when the results are rearranged by shape, a 1:2:1 ratio is
observed for color within each shape category. Likewise, when the data are
rearranged by color, a 1:2:1 ratio is observed for shape within each color
category.
9 long, red
19 oval, red
9 round, red
37 red
15 long, purple 8 round, white
32 oval, purple 16 oval, white
16 round, purple
8 long, white
63 purple
32 white
A 1:2:1 ratio is observed when there is a heterozygous × heterozygous cross.
Therefore, the original cross was a dihybrid cross. Both oval and purple must
represent an incomplete dominant phenotype.
Let L = long, L' = round, R = red and R' = white. The cross becomes
P
L/L ; R'/R' × L'/L' ; R/R
Chapter Six 109
F1
F2
L/L' ; R/R' × L/L' ; R/R'
1/4
L/L ×
1/4 R/R =
1/16
1/2
1/8
1/4
1/2
L/L' ×
L'/L' ×
1/8
1/2
1/4
1/8
1/16
1/2
1/8
R/R' =
R'/R' =
1/16
oval, red
oval, purple
oval, white
round, red
round, purple
round, white
A long, purple × oval, purple cross is as follows
P
F1
L/L ; R/R' × L/L' ; R/R' '
1/2
L/L ×
1/2
10.
R/R' =
R'/R' =
1/4 R/R =
1/4
b.
1/16
1/4 R/R =
1/4
1/4
R/R' =
R'/R'=
long, red
long, purple
long, white
L/L' ×
1/4 R/R =
1/8 long, red
1/2
1/4 long, purple
R/R' =
1/4 R'/R' =
1/4 R/R =
1/2 R/R' =
1/4 R'/R' =
1/8 long, white
1/8 oval, red
1/4 oval, purple
1/8 oval, white
From the cross c+/cch × cch/ch the progeny are
1/4
1/4
1/4
1/4
c+/cch
full color
+
h
c /c
full color
ch
ch
c /c chinchilla
cch/ch
chinchilla
Thus, 50 percent of the progeny will be chinchilla.
11.
a.
The data indicate that there is a single gene with multiple alleles. All the ratios
produced are 3:1 (complete dominance), 1:2:1 (incomplete of codominance),
or 1:1 (test cross). The order of dominance is
black > sepia > cream > albino
Cross
Parents
Cross 1: b/a × b/a
Cross 2: b/s × a/a
Progeny
3 b/– : 1 a/a
1 b/a : 1 s/a
Cross 3: c/a × c/a
3 c/– : 1 a/a
Conclusion
black is dominant to albino.
black is dominant to sepia;
sepia is dominant to albino.
cream is dominant to albino.
110 Chapter Six
Cross 4: s/a
Cross 5: b/c
Cross 6: b/s
Cross 7: b/s
Cross 8: b/c
Cross 9: s/c
Cross 10: c/a
b.
12.
×
×
×
×
×
×
×
c/a
a/a
c/–
s/–
s/c
s/c
a/a
1 c/a : 2 s/– : 1 a/a
1 b/a : 1 c/a
1 b/– : 1 s/–
1 b/s : 1 s/–
1 s/c : 2 b/– : 1 c/c
3 s/– : 1 c/c
1 c/a : 1 a/a
sepia is dominant to cream.
black is dominant to cream.
“–” can be c or a.
“–” can be s, c, or a.
The progeny of the cross b/s × b/c will be 3/4 black (1/4 b/b, 1/4 b/c, 1/4 b/s) :
1/4 sepia (s/c).
Both codominance (=) and classical dominance (>) are present in the multiple
allelic series for blood type: A = B, A > O, B > O.
Parents’ phenotype
a. AB × O
b. A × O
c. A × AB
d. O × O
Parents’ possible genotypes
A/B × O/O
A/A or A/O × O/O
A/A or A/O × A/B
O/O × O/O
Parents’ possible children
A/O, B/O
A/O, O/O
A/A, A/B, A/O, B/O
O/O
The possible genotypes of the children are
Phenotype
O
A
B
AB
Possible genotypes
O/O
A/A, A/O
B/B, B/O
A/B
Using the assumption that each set of parents had one child, the following
combinations are the only ones that will work as a solution.
Parents
a. AB × O
b. A × O
c. A × AB
d. O × O
Child
B
A
AB
O
14.
The key to solving this problem is in the statement that breeders cannot develop a
pure–breeding stock and that a cross of two platinum foxes results in some
normal progeny. Platinum must be dominant to normal color and heterozygous
(A/a). An 82:38 ratio is very close to 2:1. Because a 1:2:1 ratio is expected in a
heterozygous cross, one genotype is nonviable. It must be the A/A, homozygous
platinum, genotype that is nonviable, because the homozygous recessive genotype
is normal color (a/a). Therefore, the platinum allele is a pleiotropic allele that
governs coat color in the heterozygous state and is lethal when homozygous.
16.
a.
The sex ratio is expected to be 1:1.
b.
The female parent was heterozygous for an X-linked recessive lethal allele.
This would result in 50 percent fewer males than females.
Chapter Six 111
c.
18.
Half of the female progeny should be heterozygous for the lethal allele and
half should be homozygous for the nonlethal allele. Individually mate the F1
females and determine the sex ratio of their progeny.
In order to do this problem, you should first restate the information provided. The
following two genes are independently assorting
h/h = hairy
H/h = hairless
H/H = lethal
a.
s/s = no effect
S/s suppresses H/h, giving hairy
S/S = lethal
The cross is H/h ; S/s × H/h ; S/s. Because this is a typical dihybrid cross, the
expected ratio is 9:3:3:1. However, the problem cannot be worked in this
simple fashion because of the epistatic relationship of these two genes.
Therefore, the following approach should be used.
For the H gene, you expect 1/4 H/H : 1/2 H/h : 1/4 h/h. For the S gene, you
expect 1/4 S/S : 1/2 S/s : 1/4 s/s. To get the final ratios, multiply the frequency
of the first genotype by the frequency of the second genotype.
1/4
H/H
all progeny die regardless of the S gene
1/4
1/2
H/h
×
S/S =
1/2 S/s =
1/4 s/s =
1/8 H/h ; S/S
1/4
1/8
H/h ; S/s
H/h ; s/s
die
hairy
hairless
1/4
1/16 h/h ; S/S die
S/S =
1/4 h/h
1/2 S/s =
1/8 h/h ; S/s
×
hairy
1/4 s/s =
1/16 h/h ; s/s hairy
Of the 9/16 living progeny, the ratio of hairy to hairless is 7:2.
b.
This cross is H/h ; s/s × H/h ; S/s. A 1:2:1 ratio is expected for the H gene
and a 1:1 ratio is expected for the S gene.
1/4
H/H
1/2
H/h
1/4 h/h
all progeny die regardless of the S gene
×
×
1/2
S/s =
1/2 s/s =
1/4
1/2
1/8
S/s =
1/2 s/s =
1/4
1/8
H/h ; S/s
H/h ; s/s
hairy
hairless
h/h ; S/s
h/h ; s/s
hairy
hairy
Of the 3/4 living progeny, the ratio of hairy to hairless is 2:1.
19.
a.
The mutations are in two different genes as the heterokaryon is prototrophic
(the two mutations complemented each other).
b.
leu1+ ; leu 2– and leu1– ; leu2+
112 Chapter Six
c.
With independent assortment, expect
1/4
1/4
1/4
1/4
21.
23.
a.
colorless
b.
magenta
c.
colorless
d.
p/p ; Q/Q
P/P ; q/q
p/p ; q/q
e.
9 red : 4 colorless : 3 magenta
a.
This is an example where one phenotype in the parents gives rise to three
phenotypes in the offspring. The “frizzle” is the heterozygous phenotype
and shows incomplete dominance.
b.
25.
leu1+ ; leu 2–
leu1– ; leu2+
leu1– ; leu 2–
leu1+ ; leu 2+
P
A/a (frizzle) × A/a (frizzle)
F1
1 A/A (normal) : 2 A/a (frizzle) : 1 a/a (woolly)
If A/A (normal ) is crossed to a/a (woolly), all offspring will be A/a (frizzle).
It is possible to produce black offspring from two pure-breeding recessive albino
parents if albinism results from mutations in two different genes. If the cross is
designated
A/A ; b/b × a/a ; B/B
all offspring would be
A/a ; B/b
and they would have a black phenotype because of complementation.
28.
It is possible to produce normally pigmented offspring from albino parents if
albinism results from mutations in either of two different genes. If the cross is
designated
A/A · b/b × a/a · B/B
then all the offspring would be
A/a · B/b
and they would have a pigmented phenotype because of complementation.
Chapter Six 113
32.
To solve this problem, first restate the information.
A/–
R/–
yellow
black
A/– ; R/–
a/a ; r/r
gray
white
The cross is gray × yellow, or A/– ; R/– × A/– ; r/r. The F1 progeny are
3/8 yellow
1/8 black
3/8 gray
1/8 white
For white progeny, both parents must carry an r and an a allele. Now the cross
can be rewritten as: A/a ; R/r × A/a ; r/r
35.
The results indicate that two genes are involved (modified 9:3:3:1 ratio), with white
blocking the expression of color by the other gene. The ratio of white : color is
3:1, indicating that the F1 is heterozygous (W/w). Among colored dogs, the ratio is
3 black : 1 brown, indicating that black is dominant to brown and the F1 is
heterozygous (B/b). The original brown dog is w/w ; b/b and the original white
dog is W/W ; B/B. The F1 progeny are W/w ; B/b and the F2 progeny are
9 W/– ; B/–
3 w/w ; B/–
3 W/– ; b/b
1 w/w ; b/b
white
black
white
brown
39.
Pedigrees like this are quite common. They indicate lack of penetrance due to
epistasis or environmental effects. Individual A must have the dominant autosomal
gene, even though she does not express the trait, as both her children are affected.
44.
a.
b.
The genotypes are
P
B/B ; i/i × b/b ; I/I
F1
B/b ; I/ihairless
F2
9 B/– ; I/–
3 B/– ; i/i
3 b/b ; I/–
1 b/b ; i/i
hairless
straight
hairless
bent
In order to solve this problem, first write as much as you can of the progeny
genotypes.
4 hairless
3 straight
1 bent
–/– ; I/–
B/– ; i/i
b/b ; i/i
Each parent must have a b allele and an i allele. The partial genotypes of the
parents are
114 Chapter Six
–/b ; –/i × –/b ; –/i
At least one parent carries the B allele, and at least one parent carries the I
allele. Assume for a moment that the first parent carries both. The partial
genotypes become
B/b ; I/i × –/b ; –/i
Note that 1/2 the progeny are hairless. This must come from a I/i × i/i cross.
Of those progeny with hair, the ratio is 3:1, which must come from a B/b ×
B/b cross. The final genotypes are therefore
B/b ; I/i × B/b ; i/i
51.
Note that the F2 are in an approximate 9:6:1 ratio. This suggests a dihybrid cross
in which A/– ; b/b has the same appearance as a/a ; B/–. Let the disc phenotype be
the result of A/– ; B/– and the long phenotype be the result of a/a ; b/b. The
crosses are
P
A/A ; B/B (disc) × a/a ; b/b (long)
F1
A/a ; B/b (disc)
F2
9
3
3
1
A/– ; B/– (disc)
a/a ; B/– (sphere)
A/– ; b/b (sphere)
a/a ; b/b (long)