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Year 13 Biology
GENETICS PROBLEMS
1.
In mice, black coat is dominant to brown coat.
a) What offspring would result from a heterozygous black mouse being crossed with a brown mouse?
b) Give details of how you would find out the genotype of a black mouse.
2.
Grey seed coat colour in peas is dominant to white and tall stem is dominant to short.
a) If pure breeding tall plants which bear white seeds are crossed with pure breeding short plants which bear grey
seeds, what will be the phenotype of the F1?
b) If the F1 plants are selfed, what will be the expected phenotypes on the F2 generation and in what ratios are
they likely to occur?
3.
A pure breeding red flowered antirrhinum was crossed with a pure bred white variety. All the F1 had pink flowers.
a) How many kinds of pollen would be made by a
i) red flower
ii) pink flower
b) Using the symbols R and r for red and white respectively, show what the offspring would be if the F1 plants
were inbred.
c) What phenomenon is illustrated in this example?
4.
In cattle, red coat (R) is co-dominant to white (r) with the heterozygotes being roan in colour. Hornless (H) is
dominant to horned (h). A bull, heterozygous for both characters was mated with a roan, horned cow. What would be
the probability of the resulting calf being roan and horned?
5.
In the Andalusian fowl, the allele for black feathers is incompletely dominant to that for white feathers and the
phenotype of the heterozygote is described as blue. The texture of the feathers is controlled by a second gene locus,
silky feathers being recessive to normal. Crosses were made amongst blue birds heterozygous at the silky locus.
What genotype ratios would be expected amongst the offspring?
6.
The four blood groups of the ABO blood group system are determined by a series of multiple alleles. The allele IO is
recessive and the alleles IA and IB are co-dominant. The table below shows the blood group phenotypes and possible
genotypes.
Blood Group Phenotypes
Group O
Group AB
Group A
Group B
a)
b)
Possible Genotypes
IOIO
IAIB
IA or IO
IB or IO
Give the possible blood groups of children of the following parents
i) Group O mother and group A father
ii) Group B mother and group A father.
In the case of disputed paternity, a woman claims that a man, whose blood is group AB, is the father of her
group O child. Could her claim be correct. Explain.
7.
In cats, a pair of alleles (B and b) are sex linked. In females the homozygous condition X BXB gives a black coat
colour, XbXb gives a ginger coat and XBXb gives a patchwork of ginger and black called tortoiseshell. A tortoiseshell
female is mated with a black male.
a) What are the genotypes of the parents and the offspring?
b) What is the probability of any given male offspring being black?
c) Why is it better to refer to probability than percentages in this situation?
8.
Haemophilia is a hereditary inability to make one of the blood clotting proteins due to an allele on the X chromosome.
Answer true or false to the following statements.
a) The condition only occurs in males.
b) A male can inherit the allele from either of his mother’s parents.
c) When both parents carry the defective allele, all the son’s are affected.
9.
Two genes govern comb shape in poultry, R and P. If R and P are both present a walnut comb is produced; if only R
a rose comb; if only P a pea comb. In the double recessive condition (rrpp) a single comb only is produced.
a) What is the name given to this type of gene interaction?
b) What would be the comb character of the offspring from the following crosses:
i)
RrPp X RrPp
ii)
RrPP X RRPp
(Please Turn Over)
c) A rose-combed fowl crossed with a walnut-combed fowl produces offspring of which 3/8 are walnut,
1/ peas and 1/ single. What are the genotypes of the parents? Explain how you arrived at
your answer.
8
8
3/ rose,
8
10. The following four genes of tomato are linked; M (normal) versus m (mottled) leaf, H (smooth versus h (hairy)
epidermis, P (purple) versus p (green) stem colour, S (simple) versus s (compound) inflorescence. Recombination
percentages are as follows;
between genes M and H = 8%
between genes M and P = 25%
between genes H and P =17%
between genes S and h = 27%
Draw a genetic map for the four genes.
11. King and Castle discovered that in rats, albino and waltzer are linked. Both mutant alleles are recessive to the wild
type. In a back cross the following progeny were obtained.
Phenotype
Albino, non waltzer
Albino, waltzer
Coloured, non-waltzer
Coloured waltzer
a)
b)
Number
146
38
37
159
Define gene symbols and, using the conventional method of showing linked genes, give the genotype of the
parents of the back cross progeny.
If the genes were not linked, how many back cross progeny would be expected in each phenotypic
class?
12. Albinism is a disease which has a pleiotropic effect. What is meant by this term?
13. Certain genes have the ability to suppress the expression of a gene at a second locus. In pumpkin, colour is
recessive to no colour at one allelic pair. This recessive allele must be expressed before the specific colour allele at
a second locus is expressed. At the first gene, white coloured squash is dominant to coloured squash, and the gene
symbols are W = white and w = coloured. At the second gene, yellow is dominant to green, and the symbols used
are G = yellow and y = green. If the dihybrid is self-fertilised what will the phenotypic ratio be?
14. Phenylketonuria (PKU) is an inherited error of metabolism caused by a deficiency in the enzyme phenylalanine
hydroxylase. Loss of this enzyme results in mental retardation, organ damage and unusual posture. It is an
autosomal recessive disorder.
a) How many affected alleles must a person have before the recessive disorder is expressed?
b) PKU can be controlled by diet showing there is some environmental influence on this disease. How is it that the
environment can have an influence on the phenotype?
15. The wings of the Drosophila (fruit fly) may be curled or semi-curled. Flies which are homozygous for curled wings
(cu/cu) and reared at high temperatures have wings that are curled up at the tips. Offspring from these flies which
are reared at low temperatures show straight wings.
With reference to the genotype and phenotype of the Drosophila, explain how the wing shape differs in relation to
temperature.
16. In the pedigrees below the gene for the shaded condition is rare in the general population. For each pedigree, say
whether the condition represented by the shaded character is autosomal recessive, autosomal dominant, X linked
recessive or X linked dominant, justifying your answer in each case.
GENETIC PROBLEMS - ANSWERS
1.
B = black b = brown
a) Bb x bb
B
b
Bb
b
bb
Offspring = ½ black, ½ brown
b) You would need to do a testcross which involves crossing the black mouse of unknown genotypes with a brown
(homozygous recessive) mouse. If the offspring were all black then the unknown was probably homozygous; if you got ½ brown
and ½ black then it must be heterozygous.
2.
G = grey
T = tall
g = white
t = short
TTgg x ttGG
b)
Selfed means self pollinated
TtGg x TtGg
F2’s
3.
9: tall and grey ‘
3: short and grey
R = red
a)
b)
c)
4.
F1’s are all TtGg i.e. all tall and grey
a)
i)
ii)
3; tall and white
1: short and white
r = white
Pollen is the male sex cell
Red flowers must be homozygous – RR. They can only produce pollen containing a red allele i.e. one type. Pink
flowers are heterozygous – Rr. They can produce pollen which contains R or r i.e. two types
Rr x Rr
R
r
R
RR
Rr
r
Rr
rr
Offspring = ¼ red, ½ pink, ¼ white
Incomplete dominance
Rrhh x Rrhh
RH
Rh
rH
rh
Rh
RRHh
RRhh
RrHh
Rrhh
rh
RrHh
Rrhh
rrHh
rrhh
BS
Bs
bS
bs
BS
BBSS
BBSs
BbSS
BbSs
Bs
BBSs
BBss
BbSs
Bbss
bS
BbSS
BbSs
bbSS
bbSs
bs
BbSs
Bbss
bbSs
bbss
Probability of Rrhh = ¼
5.
B = black b = white
S = normal s = silky
Bb = blue
BbSs x BbSs
Black and normal = 3
Blue and normal = 6
Black and silky = 1
Blue and silky = 2
White and normal = 3
White and silky = 1
(Please Turn Over)
6.
a)
i) If
IOIO x IAIA
I OI A
O or A
Or
ii) If
Or
Or
Or
IOIO x IAIO
IOIB
IBIB
IOIB
IBIB
x
x
x
x
I OI A I OI O
IAIA
IAIA
IAIO
IAIO
IOIA IAIB
IAIB
IOIA IOIO
IAIB IOIB
A
B
AB
O
IOIB IAIB
b) A child of blood group O has the genotype IOIO. One of the IO alleles must come from the father. If the
group AB he has alleles IA and IB. Therefore he cannot be the father of an O group child.
7.
a)
b)
XBXb x XBXb
XB
XB XBXB
Y
c)
father is blood
Xb
XBXb
XBY
XbY
The probability of any given male being black is 1/2
In a litter of kittens the numbers are far too small for percentages to be meaningful.
8.
a) False – Much rarer in females because she must inherit the haemophilic allele from each one of her
parents.
b) True – His mother has two X chromosomes, one inherited from each of his parents.
c)
False – Unless the mother is homozygous, half of her sons will receive her normal allele and be unaffected. The father
does not hand on his defective allele to his sons.
9.
a)
b)
c)
10.
9/
3
3
1
16 walnut, /16 rose, /16 pea, /16 single
All walnut
Rrpp X RrPp
M
H
P
S
8
25
10
17
27
11. a)
Let w be the allele for waltzer and W the wild type allele
Let a be the allele for albino and A the wild type allele
The parents of the backcross progeny had the genotypes:
A
w
a
a
b)
W
a
w
w
Total number of progeny = 380
Expect a 1:1:1:1 i.e. 95 for each type.
12. The gene causing albinism also causes other effects.
13. 12 white: 3 yellow :1 green
14. The genes do not determine exactly what the phenotype is, rather it gives a limit as to the kinds of phenotypes this organism can
have if it has this particular genotype. The environment affects it as many things essential to the organism’s normal body
processes com from the environment. E.g. food and energy. Also the environment influences phenotype for adaptive purposes
sometimes, e.g. flowering only happens I favourable environments, or trees with the tall gene will only grow in “good” conditions.
15.
The genotype or genetic potential is homozygous (pure breeding) for curled wings but the phenotype is affected by the
temperature. In high temperatures the genotype is expressed but in low temperatures the wings are not curled and the genotype
is not expressed. If the flies reared in the low temperatures were to be transferred to high temperatures the genotype could now
be expressed and the wings would curl.
16. a)
b)
c)
d)
X linked recessive
It must be recessive because III2 differs from both his parents. Suggest sex linked because only occurs in the males and
because III2 and IV3 both inherit the condition from their mothers. Unlikely to come from II1 and II5 (fathers) because the
problem says it is rare in the general population.
Autosomal recessive
Recessive because IV1 and IV2 differ from both their parents. If it were sex-linked both II2 and II3
would show it.
Probably autosomal dominant
Since the allele for the shaded condition is rare in the general population we can assume I1 to be homozygous. Since
children are of both types I2 must be heterozygous, so the allele must be dominant. All the offspring that have the condition
have at least one parent with it also.
X linked dominant
All the daughters of the affected male I2 are affected but none of the sons.