BIO 10 Take-Home Assignment #5
KEY
Fall, 2007
1.
AABbcc will produce equal proportions of ABc and Abc gametes.
AaBBCc will produce equal proportions of ABC, aBC, ABc, and aBc
gametes.
AaBbCc will produce equal proportions of ABC, abc, Abc, aBC, aBc,
AbC, abC, and ABc gametes.
2.
Both parents are Hh. Each will produce ½ H and ½ h gametes.
H
h
H
HH
Hh
h
Hh
hh
Answer: ¾ H_ (hornless); ¼ hh (horned)
3.
A child’s sex is inherited independently from whether or not her/she
is a taster.
Probability of tasters from the cross:
T
t
T
TT
Tt
t
Tt
tt
= ¾ (T_) tasters; ¼ (tt) non-tasters
Probability of having a boy is ½; Probability of having a girl is ½
Answer:
4.
Probability child is a non-taster girl = (1/4)(1/2) = 1/8
Probability child is a taster boy = (3/4)(1/2) = 3/8
The alleles at each gene locus are inherited independently (Mendel’s
Law of Independent Assortment):
EeBbWW
x
EeBbww
(1/4)(1/4)(1) = 1/16 eebbW_
eeBBWw x EebbWw
(1/2)(0)(3/4) = 0 chance of eebbW_
5a.
ssRR x SSR’R’
All offspring SsRR’ = straight-haired roans
5b.
SsRR’ x ssRR’
sR
sR’
Answer:
6.
1
1
2
2
1
1
SR
SsRR
SsRR’
SsRR
ssRR
SsRR’
ssRR’
SsR’R’
ssR’R’
sR
ssRR
ssRR’
SR’
SsRR’
SsR’R’
sR’
ssRR’
ssR’R’
straight-haired red
curly-haired red
straight-haired roan
curly-haired roan
straight-haired white
curly-haired white
First determine as much as possible about the genotypes of the
parents and the offspring from their phenotypes:
S_B_ male
x
S_bb female
Offspring include 2 S_B_, 2 S_bb, 1 ssB_, and 1 ssbb
Since there are offspring that are ss, both parents must be Ss.
Since there are offspring that are bb, the male parent must be bb.
Answer: Male parent is SsBb and the female parent is Ssbb.
7.
The cross is XcY x XcXC
X
XC
c
Xc
XcXc
XcXC
Y
XcY
XCY
Answer: 1 black female, 1 black male, 1 orange male, one tortie
female.
8.
The man must be bb since he has blue-eyes.
His wife must be Bb since her father had blue eyes and therefore had
to give her a “b” allele.
b
B
Bb
b
bb
Answer: Half the children would have blue eyes (bb).
9.
The question here is whether the man is Bb or BB.
Answer: If the man is Bb, each child has a ½ chance of having brown
eyes. The chance that all 10 children would have brown eyes is
(1/2)10 = 1 in 1,024 since each pregnancy is an independent event.
Therefore, it is highly improbable that the man is Bb, but we cannot
be completely certain. (There is a 1 in 1024 chance that he is!) After
another brown-eyed child, the chance is lowered to 1 in 2048 – but
not to zero.
10.
bb
B_
B_
bb
B_
11.
B_
bb
There is one gene controlling the trait and three phenotypes, one of
which is intermediate between the other two. This is indicative of
incomplete dominance.
No tails: TT; short tails: TT’; long tails: T’T’
If two short tailed cats are crossed:
T
T’
T
TT
TT’
T’
TT’
T’T’
Prediction: 1 no tail: 2 short-tail: 1 long tail.
This is roughly consistent with the results that were observed:
1.5 no tail: 2 short tail: 1 long tail
A larger sample size would be needed to test the hypothesis with
scientific rigor.
12.
vv,hh
x
VV,HH
vh
VH
Vv,Hh
All offspring would be Vv,Hh and would therefore have normal wings
and a hairless (normal) body.
13.
U = white belt; u = uniform coloring
F = fused hooves; f = normal hooves
uu,ff
x
UU,FF
As in question 12, all offspring will be the same. In this case, they will
all be Uu,Ff (white belt, fused hooves)
Uu,Ff x Uu,Ff
We know from Mendel’s Second Law that when two dihybrids are
mated, the offspring will fall into the following phenotypic categories:
9
3
U_F_ uniform coloring, fused hooves
uuF_ white belt, fused hooves
3
1
14.
U_ff
uuff
uniform coloring, normal hooves
white belt, normal hooves
G = green; g = striped
S = short; s = long
gg,ss x Gg,Ss
gs
Answer:
15.
1
1
1
1
GS
SsRR
Gs
ssRR
gS
SsRR’
gs
ssRR’
green, short
green, long
striped, short
striped, long
A_,rr = yellow
aa,R_ = black
A_R_ = gray
Aarr = cream
A_R_
x
A_rr
Offspring:
3/8
A_R_
3/8
A_rr
1/8
aarr
1/8
aaR_
(gray)
(yellow)
(cream)
(black)
Answer: Since there are offspring that are aa, both parents must be
Aa. Since there are offspring that are rr, the gray parent must be Rr.
Therefore, the parents are AaRr and Aarr.
16.
A_rr
x
aaR_
Offspring:
46 A_R_ (gray)
53 A_rr (yellow)
Answer: Since all the offspring have at least one A allele, which must
have come from the yellow male parent, it is highly probable that the
yellow male parent is AArr. However, there is a slight possibility that
he is Aarr and simply failed to pass this along to any of his offspring.
To be mathematically precise, there are 99 offspring, all of which
inherited the “A” allele from their father. If the father was Aa, the
probability of this would be (1/2)99 = 6.23 x 1029!! – HIGHLY
improbable indeed! On the other hand, since there are rr offspring,
the female parent must be Rr.
So the father is (almost certainly) AArr and the mother must be aaRr.
17.
C_I_
C_ii
ccIi
ccii
= white
= colored
= white
= white
Answer: The cross is Ccii (colored cock) x ccii (white female)
How do we know that the white female is ccii? The question states
that all the offspring are colored. The only way for this to happen is if
the white female parent does not have an “I” allele, and the only
possible genotype that gives a white hen that does not have an “I”
allele is ccii.
Since there are no ccii offspring (which would be white), the cock
must be Ccii.
18.
K = hearing; k = deaf
M = deaf, regardless of genotype at K locus.
kkMm
x
KKmm
Km
kM
KkMm
km
Kkmm
Answer: Half of the offspring will have the genotype KkMm and be
deaf because they have a copy of the M allele. Half the offspring will
be hearing because they have a K allele and lack an M allele.
19.
Xb is a recessive lethal allele.
Xb,XB
x
XB,Y
X
XB
b
XB
X Xb
XB XB
B
Y
Xb Y
XB Y
Note that the shaded offspring will be dead and not counted among
the living offspring of the cross.
Answer: The ratio will be 2/3 females to 1/3 males among living
offspring.
20.
Mrs. Smith: B_
Mr. Smith: A_
Shirley: OO
Mrs. Jones: A_
Mr. Jones: A_
Jane: B_
Answer: Yes. Either couple could have produced Shirley, but the
Jones’ could not have produced Jane. On the other hand, the Smith’s
could have produced Jane. Therefore, a mix-up is consistent with the
evidence.
21.
The cross data is most consistent with the H allele causing
hairlessness and also being lethal in the homozygous state (HH).
Hh (hairless) x hh (hairy) yields ½ Hh (hairless) and ½ hh (hairy).
Note that since HH is lethal, the hairless dog in this cross must be
heterozygous (or it would be dead!).
H
h
H
HH
Hh
h
Hh
hh
This is also consistent with the results when two hairless dogs are
mated (Hh x Hh). The HH offspring are dead, leaving 2/3 living
hairless and 1/3 living hairy pups.
22.
X
Xc
c
XC
XC Xc
XC Xc
Y
Xc Y
Xc Y
Answer: Since all the males inherit their single X chromosome from
their mother, and the mother must be homozygous for the mutation
since she is affected, they are all colorblind. The daughters inherit
one “bad” X from Mom and one normal X from Dad, so are all
unaffected carriers.
23.
Yes. In order for the girl to be colorblind, she would need a carrier (or
affected) mother as well as an affected father. Since her father is not
affected, this raises serious doubt to paternity. However, it should be
noted that there is a SMALL probability that he father could be an
unaffected male whose sperm underwent a spontaneous, new
mutation during its production. If this mutant sperm fertilized an egg
carrying a mutant X from the mother, the girl would be colorblind.