InheritanceAQA A-Level Biology: Topic test

20 questions, 54 marks

AQA A-Level Biology

Inheritance topic test

Total 54 marks

Name

Class

Date

1
Huntington's disease is caused by a dominant allele, $H$ . The normal allele is $h$ . A man who has Huntington's disease, and whose mother did not have the disease, has a child with a woman who does not have the disease.
(a)
What is the genotype of the man?
[1 mark]
A $HH$
B $hh$
C $Hh$
DEither $HH$ or $Hh$
(b)
What is the probability that the child has Huntington's disease?
[1 mark]
A0.50
B0.25
C0.75
D1.00
(c)
The couple have three children, none of whom has Huntington's disease. Explain why the probability that a fourth child will have the disease is still 0.5.
[2 marks]
Total for question 1: 4 marks
2
In guinea pigs, black fur ( $B$ ) is dominant to brown fur ( $b$ ) and short hair ( $S$ ) is dominant to long hair ( $s$ ). The two genes are on different chromosomes. Two guinea pigs that are heterozygous for both genes are crossed.
(a)
Which combination of alleles could be present in a single gamete from these guinea pigs?
[1 mark]
A $Bb$
B $bS$
C $BBSs$
D $Ss$
(b)
What is the expected phenotypic ratio in the offspring?
[1 mark]
A3 : 1
B1 : 1 : 1 : 1
C9 : 7
D9 : 3 : 3 : 1
(c)
The cross produces 320 offspring. Calculate the number expected to have brown fur and long hair. Show your working.
[2 marks]
Total for question 2: 4 marks
3
In sweet peas, the gene for flower colour has alleles $P$ (purple, dominant) and $p$ (red), and the gene for pollen shape has alleles $L$ (long, dominant) and $l$ (round). The two genes are on the same chromosome. A plant with one chromosome carrying $P$ and $L$ and its homologue carrying $p$ and $l$ was crossed with a plant that is $pl/pl$ . The 520 offspring were: 242 purple long, 238 red round, 22 purple round and 18 red long.
(a)
Explain why most of the offspring are purple long or red round.
[3 marks]
(b)
Explain the origin of the purple round and red long offspring, and calculate the percentage of the offspring that are recombinants.
[4 marks]
Total for question 3: 7 marks
4
In a species of garden plant, flower pigment is made in two steps. Enzyme 1, coded for by allele $A$ , converts a colourless precursor into a colourless intermediate. Enzyme 2, coded for by allele $B$ , converts the intermediate into a purple pigment. The alleles $a$ and $b$ code for non-functional enzymes. The two genes are on different chromosomes. A plant with no functional enzyme 1, or with no functional enzyme 2, has white flowers. Two purple-flowered plants that were heterozygous for both genes were crossed.
(a)
Explain, using the genotypes of the plants, why the ratio of phenotypes expected in the offspring is 9 purple : 7 white.
[6 marks]
(b)
The cross produced 280 offspring: 168 purple and 112 white. Use a chi-squared ( $\chi^2$ ) test to decide whether these results fit the ratio expected in part (a). The critical value for 1 degree of freedom at $p = 0.05$ is 3.84.
[6 marks]
Total for question 4: 12 marks
5
In cats, a gene on the X chromosome has two codominant alleles: $X^{B}$ gives black fur and $X^{O}$ gives orange fur. A heterozygous female is tortoiseshell. There is no equivalent gene on the Y chromosome. A tortoiseshell female mates with a black male.
(a)
Which kitten cannot be produced by this mating?
[1 mark]
AA black female
BAn orange male
CA tortoiseshell female
DAn orange female
(b)
What proportion of the kittens is expected to be black?
[1 mark]
A25%
B50%
C75%
D100%
(c)
Explain why tortoiseshell cats are almost always female.
[2 marks]
Total for question 5: 4 marks
6
A breeder test crosses a heterozygous black rabbit ( $Bb$ ) with a brown rabbit ( $bb$ ). He expects black and brown offspring in the ratio 1 : 1. Of 100 offspring, 58 are black and 42 are brown. He decides to use a chi-squared test.
(a)
Which statement is the null hypothesis for his test?
[1 mark]
AThere is no significant difference between the observed and expected numbers; any difference is due to chance.
BThere is a significant difference between the observed and expected numbers.
CThe black allele is dominant to the brown allele.
DAll of the offspring are heterozygous.
(b)
How many degrees of freedom should he use?
[1 mark]
A0
B2
C1
D99
(c)
Calculate the value of $\chi^2$ and state the conclusion. The critical value at $p = 0.05$ is 3.84.
[2 marks]
Total for question 6: 4 marks
7
In rabbits, coat colour is controlled by one gene with four alleles: $C$ gives full colour, $c^{ch}$ gives chinchilla, $c^{h}$ gives Himalayan and $c$ gives albino. The order of dominance is $C > c^{ch} > c^{h} > c$ .
(a)
A full-colour rabbit with genotype $Cc^{h}$ is crossed with an albino rabbit. Give the genotypes and phenotypes of the offspring and their ratio.
[3 marks]
(b)
A chinchilla rabbit is crossed with a Himalayan rabbit and the offspring include chinchilla, Himalayan and albino rabbits. Deduce the genotypes of the parents and the ratio of phenotypes expected in the offspring.
[4 marks]
Total for question 7: 7 marks
8
In maize, the gene for kernel colour has alleles $P$ (purple, dominant) and $p$ (yellow), and the gene for kernel texture has alleles $S$ (smooth, dominant) and $s$ (shrunken). A plant heterozygous for both genes was test crossed with a plant that is homozygous recessive for both genes. The 1000 offspring were: 440 purple smooth, 430 yellow shrunken, 70 purple shrunken and 60 yellow smooth. A student suggests that the two genes are on different chromosomes and assort independently.
(a)
Use a chi-squared test to test the student's suggestion. The critical value for 3 degrees of freedom at $p = 0.05$ is 7.82.
[6 marks]
(b)
Explain the results of this test cross in terms of the positions of the genes on the chromosomes, and calculate the percentage of recombinant offspring.
[6 marks]
Total for question 8: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).