Download Activity 6.8: Genetics problems—suggested

Heinemann Biology Activity Manual
Activity 6.8: Genetics problems—suggested answers
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a
Straight frontal hairline genotype = ww.
b
Widow’s peak genotype = WW or Ww.
c
Parents must both be Ww since they produce a child who is ww.
W
w
W
WW
Ww
w
Ww
ww
d
Probability of a child with widow’s peak is 0.75 (three in four).
e
Probability of a child with straight frontal hairline is 0.25 (one in four).
a
Purple (P) is dominant. If blue were dominant then crossings of blue flowers (BB or
Bb) with blue flowers (Bb) would produce some purple flowers.
b
Purple plant genotype must be Pp in order to produce blue progeny. The blue plant
can only be pp since blue is the recessive allele.
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a
If both the man and woman are normally pigmented yet each has a parent who is an
albino (aa), their other parents must have been AA or Aa. Therefore the man and
woman must both be heterozygous for the trait (Aa).
A
a
A
AA
Aa
a
Aa
aa
The probability of them having an albino child (aa) is 0.25 (one in four).
b
The man is an albino, so his genotype is aa. The woman’s father is an albino but she
is normally pigmented, so her genotype is Aa.
A
a
a
Aa
aa
a
Aa
aa
The probability of the couple having an albino child is 0.5 (one in two).
c
The man is an albino (aa) and the woman is normally pigmented with no family
history of albinism (AA).
A
A
a
Aa
Aa
a
Aa
Aa
The probability of the couple having an albino child is 0.
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a
Homozygous black crossed with white:
B
B
b
Bb
Bb
b
Bb
Bb
Genotypic ratio in F1 = all are Bb
Phenotypic ratio in F1 = all are black
b
Two heterozygous blacks:
B
b
B
BB
Bb
b
Bb
bb
Genotypic ratio of F1 = BB : Bb : bb = 1 : 2 : 1
Phenotypic ratio of F1 = 3 black : 1 white
c
Heterozygous black crossed with white:
B
b
b
Bb
bb
b
Bb
bb
Genotypic ratio of F1 = Bb : bb = 1 : 1
Phenotypic ratio of F1 = 1 black : 1 white
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a
The genotype of at least one parent must be Pp, since both purple-eyed (PP or Pp)
and yellow-eyed (pp) offspring are produced. The genotype of the other parent must
be either Pp or pp, because a PP parent would result in all offspring being purpleeyed.
b
Genotypes of offspring, assuming both parents are Pp:
P
p
P
PP
Pp
p
Pp
pp
The offspring could be PP, Pp or pp in the ratio 1 : 2 : 1. That is, the probability of
purple eyes is 0.75 and yellow eyes 0.25.
Genotypes of offspring, assuming one parent is Pp and the other is pp:
P
p
p
Pp
pp
p
Pp
pp
The offspring could be Pp or pp in the ratio 1 : 1. That is, the probability of purple
eyes is 0.5 and yellow eyes 0.5.
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a
At least one parent must have been heterozygous (Ll) because some short-winged
(ll) offspring were produced. Since there were many more long-winged than shortwinged offspring, both parents must have been Ll.
b
Genotypes of offspring:
L
l
L
LL
Ll
l
Ll
ll
Long-winged offspring may be LL or Ll in the ratio 1 : 2, so two-thirds of the longwinged offspring should be heterozygous (Ll).
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a
This is an example of codominant inheritance.
b
Phenotypes of offspring:
R
W
R
RR
RW
W
RW
WW
Phenotypic ratio of F1 = 2 roan : 1 red : 1 white
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B1 = steel blue
B2 = turquoise
Cross between a royal blue (B1B2) and a turquoise (B2B2) fish:
B1
B2
B2
B1B2
B2B2
B2
B1B2
B2B2
Phenotypic ratio of F1 = 1 royal blue : 1 turquoise
Cross between two royal blue fish:
B1
B2
B1
B1B1
B1B2
B2
B1B2
B2B2
Phenotypic ratio of F1 = 0.5 royal blue : 0.25 turquoise : 0.25 steel blue
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Cross between two people who have the sickle cell trait HbNHbS.
Phenotypes of offspring:
HbN
HbS
HbN
HbNHbN
HbNHbS
HbS
HbNHbS
HbSHbS
Phenotypic ratio of F1 = 2 sickle cell trait : 1 normal : 1 sickle cell anaemia.
10 a
All the sufferers were male.
b
This gene is sex-linked recessive.
c
Queen Elizabeth II is heterozygous (XXh).
11
There is a 0.5 (one in two) chance that a grandson will be colour-blind, because the
mothers will both be carriers.
12 Dominant : recessive = 3 : 1.
13 100% of F1 will show the codominant traits.
14 100% of F1 dominant phenotype; all females heterozygous.
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