Genetics- Inheritance Exam
Questions LP1 Name: ________________________
Class: ________________________
Date: ________________________
Time: 38 minutes
Marks: 34 marks
Page 1 of 13
Q1.
A single gene controls the presence of hair on the skin of cattle. The gene is carried on
the X chromosome. Its dominant allele causes hair to be present on the skin and its
recessive allele causes hairlessness.
The diagram shows the pattern of inheritance of these alleles in a group of cattle.
(a) Use evidence from the diagram to explain
(i) that hairlessness is caused by a recessive allele
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(2)
(ii) that hairlessness is caused by a gene on the X chromosome.
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(1)
(b) What is the probability of the next calf born to animals 5 and 6 being hairless?
Complete the genetic diagram to show how you arrived at your answer.
Phenotypes of parents Female with hair Male with hair
Genotypes of parents _______________ _______________
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Gametes _______________ _______________
Genotypes of offspring __________________________
Phenotypes of offspring __________________________
Probability of next calf being hairless __________________________
(4)
(Total 7 marks)
Q2.
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 _______________ _______________
Offspring genotypes
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Offspring phenotypes
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(Total 4 marks)
Page 3 of 13
Q3.
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 bi gives cinnamon fur.
• Allele B is dominant to both allele b and bi.
• Allele b is dominant to allele bi.
(i) Complete the table to show the phenotypes of cats with each of the genotypes
shown.
Genotype Phenotype
Bbi
bbi
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 ___________ ___________
Page 4 of 13
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)
(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)
Q4.
Duchenne muscular dystrophy is a sex-linked inherited condition which causes
degeneration of muscle tissue. It is caused by a recessive allele. The diagram shows the
inheritance of muscular dystrophy in one family.
Page 5 of 13
(a) Give evidence from the diagram which suggests that muscular dystrophy is
(i) sex-linked; _____________________________________________________
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(1)
(ii) caused by a recessive allele. ______________________________________
______________________________________________________________
(1)
(b) Using the following symbols,
XD = an X chromosome carrying the normal allele
Xd = an X chromosome carrying the allele for muscular dystrophy
Y = a Y chromosome
give all the possible genotypes of each of the following persons.
5 _________________________________________________________________
6 _________________________________________________________________
7 _________________________________________________________________
8 _________________________________________________________________
(2)
(c) A blood test shows that person 14 is a carrier of muscular dystrophy. Person 15 has
Page 6 of 13
recently married person 14 but as yet they have had no children. What is the
probability that their first child will be a male who develops muscular dystrophy?
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(1)
(Total 5 marks)
Q5.
Red-green colour blindness is caused by a mutation in the gene coding for one of the
opsin proteins which are needed for colour vision. The diagram shows the inheritance of
red-green colour blindness in one family.
Person 12 is pregnant with her fourth child. What is the probability that this child will be a
male with red-green colour blindness? Explain your answer by drawing a genetic diagram.
Use the following symbols
XR = an X chromosome carrying an allele for normal colour vision
Xr= an X chromosome carrying an allele for red-green colour blindness
Y = a Y chromosome
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Probability = ____________________
(Total 4 marks)
Q6.
Human ABO blood groups are determined by the presence or absence of two antigens (A
and B) on the plasma membrane of the red blood cells. The inheritance of these blood
groups is controlled by three alleles:
I A – determines the production of antigen A
I – determines the production of antigen B
B
I o – determines the production of no antigen
Alleles I A and I B are codominant. Allele I o is recessive to both.
The pedigree shows the pattern of inheritance of these blood groups in a family over three
generations.
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(a) (i) How many antigen-determining alleles will be present in a white blood cell?
Give a reason for your answer.
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(1)
(ii) Which antigen or antigens will be present on the plasma membranes of red
blood cells of individual 5?
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(1)
(b) If individuals 6 and 7 were to have another child, what is the probability that this
child would be male and blood group A? Explain your answer.
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(3)
(Total 5 marks)
Page 9 of 13
Mark schemes
Q1.
(a) (i) 1. Animal 2 / 5 has hair but offspring do not;
Accept parents as alternative to animals 2 and 5
2. So 2 / 5 parents must be heterozygous / carriers;
1 + 3: Allow reference to children / offspring for animals 7 + 8
OR
3. 4 / 7 / 8 are hairless but parents have hair;
Ignore reference to individuals 1 and 6
4. So 2 / 5 must be heterozygous / carriers;
2
(ii) Hairless males have fathers with hair / 4 is hairless but 1 is hairy / 7 and /
or 8 are hairless but 6 is hairy / only males are hairless;
Ignore references to other individuals
Ignore reference to genotypes
Allow credit for candidate who states that evidence is not
conclusive / pedigree possible with autosomal character;
1
(b) 1. Parental genotypes
XHXh and XHY
Gametes
XH Xh XH Y;
Accept any letter for gene but capital letter must represent
dominant allele.
Both parental genotypes and gametes must be correct
2. Genotypes of offspring
XHXH, XHY, XHXh, XhY;
Allow for offspring genotypes correctly derived from gametes
given by candidate;
3. Phenotypes of offspring
female with hair
male with hair
male hairless;
Allow phenotypes correctly derived from offspring genotype
Allow H ≡ X H, h ≡ Xh
4. 0.25 / ¼ / 1 in 4 / 25 %
Ignore 1:3 in context of correct probability
Reject 1:4
4
[7]
Page 10 of 13
Q2.
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]
Q3.
(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
(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;
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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]
Q4.
(a) (i) Only seen in males / not in females;
1
(ii) Unaffected parents / mother → child with M.D. /
(1 ×)2 → 5 / (3 ×) 4 → 11 / 8 (× 9) → 13;
1
(b) 5 = X dY
6 = X DY
7 = XDXd AND XDXD
8 = X X ;;
D d
All 4 correct = 2 marks
2 or 3 correct = 1 mark
max 2
(c) ¼ / 0.25 / 25% / 1:3 / 1 in 4; (NOT ‘1:4’)
1
[5]
Page 12 of 13
Q5.
parental genotypes correct: XRXr AND XRY;
gametes correct for candidate’s parental genotypes;
offspring genotypes correct and colourblind male identified as XrY /
correct genotypes derived from cand’s gametes and identify XrY;
correct probability = ¼ / 0.25 / 25% / 1 in 4 / 1:3 ;
[4]
Q6.
(a) (i) Two, as white blood cells are diploid cells / alleles are present on each
chromosome of an homologous pair / one maternal and one paternal;
1
(ii) A and B
(reject IA and IB)
1
(b) 1 in 8 / 1 / 8 / 12.5% / 1:7 / 0.125;
(Reject 1:8) parents IAIO and IBIO ;
give 1:3 / ¼ / 1 in 4 / 25% probability of blood group A and half will be male;
(accept 2nd and 3rd points from a suitable genetic diagram)
3
[5]
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