Dubai English Speaking College
Q1. The diagram shows the inheritance of cystic fibrosis in one family.
(a) Cystic fibrosis is caused by a recessive allele.
Explain the evidence for this given in the diagram.
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(2)
(b) Couple 7 and 8 decide to have another child.What is the probability that this child
will be a girl with cystic fibrosis?Complete the genetic diagram to explain your
answer.Use the symbols N for the dominant allele and n for the recessive allele.
7 8
Parental phenotypes Unaffected Unaffected
Parental genotypes .............. ..............
Genotypes of gametes .................. ..................
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Offspring genotypes ..............................................................
Offspring phenotypes ..............................................................
Probability of girl with
cystic fibrosis ..............................................................
(4)
(Total 6 marks)
Q2. A breeder crossed a black male cat with a black female cat on a number of occasions.
The female cat produced 8 black kittens and 4 white kittens.
(a) (i) Explain the evidence that the allele for white fur is recessive.
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(1)
(ii) Predict the likely ratio of colours of kittens born to a cross between this black
male and a white female.
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(1)
(b) The gene controlling coat colour has three alleles. The allele B gives black fur, the
allele b gives chocolate fur and the allele b gives cinnamon fur. i
• Allele B is dominant to both allele b and b . i
• Allele b is dominant to allele b . i
(i) Complete the table to show the phenotypes of cats with each of the genotypes
shown.
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Genotype Phenotype
Bb i
bb i
Bb
(1)
(ii) A chocolate male was crossed several times with a black female.
They produced
• 11 black kittens
• 2 chocolate kittens
• 5 cinnamon kittens.
Using the symbols in part (b), complete the genetic diagram to show the
results of this cross.
Parental phenotypes Chocolate male Black female
Parental genotypes ....................... .......................
Gametes ....................... .......................
Offspring genotypes ............... ............... .................
Offspring phenotypes Black Chocolate Cinnamon
(3)
(iii) The breeder had expected equal numbers of chocolate and cinnamon kittens
from the cross between the chocolate male and black female. Explain why the
actual numbers were different from those expected.
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(1)
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(iv) The breeder wanted to produce a population of cats that would all have
chocolate fur. Is this possible? Explain your answer.
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(2)
(Total 9 marks)
Q3.Researchers investigated some characteristics of people from different parts of England. In
the north of England they selected 200 people and recorded their phenotypes for three
different characteristics.
Their results are shown in the figure below.
Phenotype produced Number of Phenotype produced Number of
by dominant allele people by recessive allele people
Tongue roller 131 Non-tongue roller 58
Right-handed 182 Left-handed 14
Straight thumb 142 Hitch-hiker thumb 50
(a) Calculate the ratio of straight thumb to hitch-hiker thumb in this study.
Ratio = .......................................................................
(1)
(b) The numbers for the tongue rolling and thumb characteristics do not add up to 200.
For each characteristic suggest one reason why the numbers do not add up to 200.
Tongue rolling ................................................................................................
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Thumb ............................................................................................................
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(2)
(c) One student looked at the researchers’ results and concluded that 91% of people in
the UK are right-handed.
Do you agree with this conclusion? Give reasons for your answer.
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(2)
(Total 5 marks)
Q4. In fruit flies, the allele for grey body, G, is dominant to the allele for ebony body, g,
and the allele for normal wings, N, is dominant to the allele for vestigial wings, n. Vestigial-
winged flies, heterozygous for grey body colour, were crossed with ebony-bodied flies,
heterozygous for normal wings.
Complete the genetic diagram to show the genotypes and phenotypes in this cross.
Parental phenotypes Grey body, vestigial wings Ebony body, normal wings
Parental genotypes .............................. ...............................
Gamete genotypes .............................. ...............................
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Offspring genotypes ..............................................................................................
Offspring phenotypes ............................................................................................
(Total 4 marks)
Q5.One form of baldness in humans is controlled by two alleles, B and b, of a single gene. This
gene is not on the X chromosome but the expression of the gene is affected by the sex of
a person.
Men who are BB or Bb will become bald. Men who are bb will not become bald.
Women who are BB will become bald. Women who are Bb or bb will not become bald.
One type of colour blindness is controlled by a sex-linked gene, found on the X
chromosome. The dominant allele XA leads to normal colour vision and the recessive
allele Xa leads to colour blindness.
(a) (i) Give all the possible genotypes of a bald man who has normal colour vision.
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(1)
(ii) Give all the possible genotypes of a woman who will not become bald and who
carries one allele for colour blindness.
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(1)
(b) A mother and a father are both heterozygous for the gene for baldness. The father
has normal colour vision and the mother is heterozygous for the gene for colour
blindness. Complete the genetic diagram to show the probability of a son of this
couple being colour blind but not becoming bald.
Father Mother
Genotypes of parents
Gametes
Genotypes of sons
Probability of son being colour blind but not becoming bald ..........................
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(4)
(Total 6 marks)
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M1. (a) Parents without CF → offspring with CF / 1 + 2 → 6 / 7 + 8 → I0;
Each parent must have CF allele / offspring receives CF allele
from both parents / both parents heterozygous / both carriers;
2
(b) Nn and Nn (no mark since awarded in (a) already)
Accept alternative symbols
N n and N n;
Ignore X and Y
NN and Nn and Nn and nn;
Correct allocation of phenotypes to genotypes;
Probability = 0.125;
Accept answers expressed as chance rather than
probability, eg 1 in 8 / 1 to 7 / 12.5%;
4
[6]
M2. (a) (i) 1. Parents are heterozygous;
Accept carriers / carries white allele
2. Kittens receive white allele from parents / black cat;
1 max
(ii) 1:1;
Answer must be expressed as a ratio that could be reduced
to 1 : 1
1
(b) (i) Black,
Chocolate,
Black;
All three correct for the mark
1
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(ii) Parental phenotypes Chocolate male Black female
1. Parental genotypes bbi Bbi;
Both genotypes needed for the mark.
1
2. Parental gametes b bi B bi ;
Allow credit if gametes are correctly derived from candidate’s
incorrect parental genotypes.
1
3. Offspring genotypes Bb, Bbi bbi bibi;
Genotype(s) must be with correct phenotype.
Allow credit if symbols other than B / b / bi have been used
correctly.
Ignore genetic diagrams unless clearly annotated.
1
Offspring phenotypes Black Chocolate cinnamon;
(iii) 1. Offspring ratios are a probability / not fixed / arise by chance /
2. gametes may not be produced in equal numbers /
3. fertilisation / fusion of gametes is random /
4. small sample;
1
(iv) 1. Possible if parents homozygous / bb;
2. Don’t know genotype of chocolate cat / chocolate cat could be
homo- or heterozygous / chocolate cat could be bb or bbi;
3. Two chocolate cats could give cinnamon kittens;
2 max
[9]
M3.(a) 2.84:1;
Accept ‘2.84 to 1’ or (just) 2.84
Do not accept 1:2.84 or 142:50
1
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(b) 1. Some embarrassed / some not willing to show tongue / cannot tell;
2. Could not decide whether thumb was straight or not / thumb bending is
judgemental / subjective;
2
(c) 1. (No) - should be 92.9% / should be calculated from 182 out of 196 /
should not be calculated from 182 out of 200;
Allow either no or yes approach but no mark awarded for no
or yes on its own
2. (Yes) – assumes 4 out of 200 use either hand;
Accept ambidextrous
3. (But) sample may not be representative;
This could be expressed in other ways e.g. only based on
one part of the country / might not be the same in different
parts of the UK / might not be representative of UK
4. Small sample size / only sampled 200;
2 max
[5]
M4. Parental genotypes: Gg nn gg Nn ;
Gamete genotypes Gn gn gN gn ;
gN gn
Gn Gg Nn Gg nn
Grey, normal Grey, vestigial
gn gg Nn gg nn
Ebony, normal Ebony, vestigial
All offspring genotypes correct;
All offspring genotypes correctly derived;
[4]
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M5.(a) (i) BBXAY, BbXAY;
1
(ii) BbXAXa, bbXAXa;
1
(b) parental genotypes − BbXAY x BbXA Xa;
1
Gametes − (BXA, bXA,) BY, bY, BXA, B Xa, bXA, b Xa ;
1
Genotypes of sons- ;
Male gametes
BY bY
BXA BBXAY BbXAY
B Xa BB Xa Y Bb Xa Y
Female
gametes
bXA BbXAY bbXAY
b Xa Bb Xa Y bb Xa Y
0.125 / 12.5% / 1/8 ;
1
[6]
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