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A Level H2 Biology Genetics Inheritance Quiz

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A-Level Biology H2 Quiz - Genetics Inheritance

Name: __________________________
Class: __________________________
Date: __________________________
Score: _______ / 45

Duration: 60 minutes
Total Marks: 45
Instructions:

Answer all questions.
Write your answers in the spaces provided.
Use black or blue ink. Diagrams should be drawn in pencil.
Marks are indicated in brackets [ ] at the end of each question or part question.

Section A: Multiple Choice and Short Structured Questions (Questions 1–5)

1. In a species of snapdragon, flower colour is determined by a single gene with two alleles, $C^R$ (red) and $C^W$ (white). The alleles show codominance. A plant with pink flowers is crossed with a plant with white flowers.
What is the expected phenotypic ratio of the offspring?
A. 1 Red : 1 White
B. 1 Red : 2 Pink : 1 White
C. 1 Pink : 1 White
D. All Pink

Answer: _______________ [1]

2. Haemophilia is a sex-linked recessive condition caused by an allele on the X chromosome. A woman who is a carrier for haemophilia marries a man who does not have the condition.
What is the probability that their first child will be a son with haemophilia?
A. 0%
B. 25%
C. 50%
D. 100%

Answer: _______________ [1]

3. The gene for cystic fibrosis is located on chromosome 7. Two healthy parents have a child with cystic fibrosis. They are expecting a second child.
What is the probability that the second child will be a healthy carrier of the cystic fibrosis allele?
A. 0.25
B. 0.50
C. 0.67
D. 0.75

Answer: _______________ [1]

4. Fig. 4.1 shows the results of a gel electrophoresis analysis for a gene associated with a genetic disorder. The gene has two alleles, $A$ and $a$ . Allele $A$ contains a restriction site that allele $a$ lacks.

(Imagine Fig 4.1: Lane 1 shows two bands. Lane 2 shows one band at the top. Lane 3 shows one band at the bottom.)

Individual X is known to be heterozygous ( $Aa$ ). Which lane represents Individual X?
A. Lane 1
B. Lane 2
C. Lane 3
D. None of the above

Answer: _______________ [1]

5. In mice, coat colour is controlled by two genes. Gene B controls pigment production ( $B$ = black, $b$ = brown). Gene C controls pigment deposition ( $C$ = colour deposited, $c$ = no colour deposited/albino). The $c$ allele is epistatic to gene B.
Two mice with genotype $BbCc$ are crossed.
What proportion of the offspring will be albino?
A. 1/16
B. 3/16
C. 4/16
D. 9/16

Answer: _______________ [1]

Section B: Structured Questions (Questions 6–15)

6. A gene in humans codes for the production of insulin. This gene exists in multiple forms within the population, some of which result in slightly different amino acid sequences but still functional insulin.
(a) Define the term allele.

_________________________________________________________________________ [1]

(b) Explain why these different forms of the insulin gene are considered alleles of the same gene rather than different genes.

_________________________________________________________________________ [2]

7. In a certain plant species, leaf shape is controlled by a single gene with two alleles. The allele for serrated leaves ( $S$ ) is dominant to the allele for smooth leaves ( $s$ ).
A heterozygous plant is self-pollinated.
(a) Construct a genetic diagram to show the cross. Use $S$ and $s$ to represent the alleles.

Parental phenotypes: _______________ x _______________
Parental genotypes: _______________ x _______________
Gametes: _______________ x _______________

Offspring genotypes:

Offspring phenotypes and ratio:
_________________________________________________________________________ [4]

(b) If 200 seeds from this cross are germinated, calculate the expected number of plants with smooth leaves. Show your working.

Working:

Answer: _______________ [1]

8. Fig. 8.1 shows the pedigree of a family affected by a rare genetic condition.

(Imagine Fig 8.1: Generation I has unaffected male and female. Generation II has two unaffected sons and one affected daughter. Generation III has children from the unaffected sons, all unaffected.)

(a) Deduce, with reasons, whether the condition is caused by a dominant or recessive allele.

_________________________________________________________________________ [2]

(b) Deduce, with reasons, whether the gene is located on an autosome or the X chromosome.

_________________________________________________________________________ [2]

(c) Individual II-3 (the affected daughter) marries a man who is homozygous normal for this gene. Calculate the probability that their first child will be affected.

_________________________________________________________________________ [1]

9. In cats, the gene for coat colour is located on the X chromosome. The allele $X^B$ codes for black fur and $X^O$ codes for orange fur. These alleles are codominant. Female cats heterozygous for these alleles ( $X^B X^O$ ) have tortoiseshell coats. Male cats are either black ( $X^B Y$ ) or orange ( $X^O Y$ ).
(a) Explain why male cats cannot have a tortoiseshell coat.

_________________________________________________________________________ [2]

(b) A black female cat is crossed with an orange male cat.
Construct a genetic diagram to determine the phenotypes of the offspring.

Parental genotypes: _______________ x _______________
Gametes: _______________ x _______________

Offspring genotypes:

Offspring phenotypes:
_________________________________________________________________________ [4]

10. A dihybrid cross was performed in Drosophila melanogaster (fruit flies). The genes for body colour (grey $G$ is dominant to ebony $g$ ) and wing length (long $L$ is dominant to vestigial $l$ ) are located on different autosomes.
Two flies heterozygous for both traits ( $GgLl$ ) were crossed.
(a) State the expected phenotypic ratio of the offspring.
_________________________________________________________________________ [1]

(b) The observed results from the cross were:

Grey body, long wing: 548
Grey body, vestigial wing: 185
Ebony body, long wing: 192
Ebony body, vestigial wing: 65

Total offspring = 990.

Complete Table 10.1 to calculate the chi-squared ( $\chi^2$ ) value.

Table 10.1

Phenotype	Observed (O)	Expected Ratio	Expected (E)	$O - E$	$(O - E)^2$	$\frac{(O - E)^2}{E}$
Grey, Long	548	9
Grey, Vestigial	185	3
Ebony, Long	192	3
Ebony, Vestigial	65	1
Total	990	16	990			$\chi^2 =$

[4]

(c) The critical value for $\chi^2$ at $p=0.05$ with 3 degrees of freedom is 7.82.
State the null hypothesis for this test.

_________________________________________________________________________ [1]

(d) Based on your calculated $\chi^2$ value and the critical value, state and explain your conclusion.

_________________________________________________________________________ [2]

11. Sickle cell anaemia is caused by a mutation in the $\beta$ -globin gene. The normal allele is $Hb^A$ and the sickle cell allele is $Hb^S$ .
(a) Describe the molecular basis of the difference between the $Hb^A$ and $Hb^S$ alleles.

_________________________________________________________________________ [2]

(b) In regions where malaria is endemic, the frequency of the $Hb^S$ allele is higher than in non-malarial regions. Explain this observation in terms of natural selection.

_________________________________________________________________________ [3]

12. Fig. 12.1 shows the structure of a nucleosome.

(Imagine Fig 12.1: DNA wrapped around histone proteins.)

(a) Identify the components labelled A and B.
A: __________________________
B: __________________________ [2]

(b) Explain how the structure of chromatin changes during the cell cycle to allow for gene expression.

_________________________________________________________________________ [2]

13. In a population of 10,000 individuals, 900 are affected by a recessive genetic disorder. Assume the population is in Hardy-Weinberg equilibrium.
(a) Calculate the frequency of the recessive allele ( $q$ ). Show your working.

Working:

Answer: _______________ [2]

(b) Calculate the number of heterozygous carriers in this population. Show your working.

Working:

Answer: _______________ [2]

14. Polydactyly (extra fingers or toes) is caused by a dominant allele ( $P$ ). However, not all individuals with the genotype $Pp$ or $PP$ exhibit the phenotype.
(a) Define the term penetrance.

_________________________________________________________________________ [1]

(b) Suggest how environmental factors or other genes might influence the expression of polydactyly.

_________________________________________________________________________ [2]

15. A student investigated the inheritance of eye colour in fruit flies. The gene is sex-linked. Red eye ( $R$ ) is dominant to white eye ( $r$ ).
A white-eyed female was crossed with a red-eyed male.
(a) Predict the phenotypes of the F1 generation.

_________________________________________________________________________ [2]

(b) The F1 females were then crossed with the F1 males. Predict the phenotypic ratio of the F2 generation.

_________________________________________________________________________ [2]

Section C: Data Response and Extended Answer (Questions 16–20)

16. Fig. 16.1 shows the results of an experiment investigating the effect of temperature on the activity of an enzyme involved in melanin production in mice.

(Imagine Fig 16.1: A graph showing enzyme activity peaking at 35°C and dropping sharply at 40°C and above. Another graph showing fur colour intensity vs temperature.)

(a) Describe the relationship between temperature and enzyme activity shown in Fig. 16.1.

_________________________________________________________________________ [2]

(b) Explain why the fur of Himalayan rabbits is white on the body but black on the ears, nose, and feet, referencing the enzyme activity.

_________________________________________________________________________ [3]

17. Cystic fibrosis (CF) is caused by mutations in the CFTR gene. Over 2,000 different mutations have been identified.
(a) Explain why individuals with different mutations in the CFTR gene may show varying severity of symptoms.

_________________________________________________________________________ [3]

(b) Gene therapy is being developed to treat CF. Describe one method of delivering a functional CFTR gene to lung cells and discuss one ethical concern associated with somatic gene therapy.

_________________________________________________________________________ [4]

18. In a species of flower, petal colour is controlled by two unlinked genes, A and B.

Gene A controls pigment production: $A$ (pigment) is dominant to $a$ (no pigment/white).
Gene B controls pigment colour: $B$ (purple) is dominant to $b$ (red).
If genotype is $aa$ , the flower is white regardless of gene B.

(a) What is this type of gene interaction called?
_________________________________________________________________________ [1]

(b) A plant with genotype $AaBb$ is self-pollinated.
Determine the phenotypic ratio of the offspring. Show your working using a Punnett square or logical deduction.

Working:

Phenotypic Ratio:
_________________________________________________________________________ [4]

19. Mitochondrial DNA (mtDNA) is inherited maternally.
(a) Explain why mtDNA is inherited only from the mother.

_________________________________________________________________________ [2]

(b) A woman with a mitochondrial disease has four children. Her husband does not have the disease.
What is the probability that each of her children will inherit the mitochondrial disease? Explain your answer.

_________________________________________________________________________ [3]

20. Discuss the advantages and disadvantages of using model organisms, such as Drosophila melanogaster or Caenorhabditis elegans, in genetic research compared to studying humans directly.
In your answer, consider:

Ethical considerations
Generation time
Genetic similarity to humans
Control of environmental variables

_________________________________________________________________________ [6]

End of Quiz

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Answers

A-Level Biology H2 Quiz - Genetics Inheritance (Answer Key)

1. C
Reasoning: Parent 1 (Pink) is $C^R C^W$ . Parent 2 (White) is $C^W C^W$ . Offspring: $C^R C^W$ (Pink) and $C^W C^W$ (White). Ratio 1:1. [1]

2. B
Reasoning: Mother $X^H X^h$ , Father $X^H Y$ . Sons receive Y from father. 50% chance son gets $X^h$ from mother. Total probability for "son with haemophilia" among all children is 1/4 (25%). [1]

3. B
Reasoning: Parents are carriers ( $Cc$ ). Cross $Cc \times Cc$ . Genotypes: 1 $CC$ : 2 $Cc$ : 1 $cc$ . Healthy carriers are $Cc$ . Probability is 2/4 = 0.50. [1]

4. A
Reasoning: Heterozygote has both alleles. Allele A cuts (smaller fragment), Allele a does not cut (larger fragment). Lane 1 shows two bands, representing both fragments. [1]

5. C
Reasoning: Epistasis. $cc$ is albino. Cross $BbCc \times BbCc$ . Probability of $cc$ is 1/4. Regardless of B/b, if $cc$ , phenotype is albino. So 1/4 (4/16) are albino. [1]

6.
(a) An alternative form of a gene. [1]
(b) They occupy the same locus [1] on the same chromosome [1] (or homologous chromosomes). [2]

7.
(a)
Parental phenotypes: Serrated x Serrated
Parental genotypes: $Ss$ x $Ss$
Gametes: $S, s$ x $S, s$
Offspring genotypes: $SS, Ss, Ss, ss$ (or 1 $SS$ : 2 $Ss$ : 1 $ss$ )
Offspring phenotypes: 3 Serrated : 1 Smooth [4]
(b)
Expected smooth ( $ss$ ) = 1/4 of total.
$200 \times 0.25 = 50$ .
Answer: 50 [1]

8.
(a) Recessive. [1] Unaffected parents (I-1, I-2) have an affected child (II-3). If dominant, one parent must be affected. [1] [2]
(b) Autosomal. [1] If X-linked recessive, the affected daughter (II-3) would have genotype $X^a X^a$ . She must receive one $X^a$ from her father (I-1). This would make the father affected ( $X^a Y$ ). But the father is unaffected. Therefore, it is autosomal. [1] [2]
(c) Individual II-3 is $aa$ . Husband is $AA$ . Offspring all $Aa$ . Probability of affected ( $aa$ ) is 0. [1]

9.
(a) Males have only one X chromosome ( $XY$ ). [1] They can only carry one allele ( $X^B$ or $X^O$ ), not both. Codominance requires both alleles to be present. [1] [2]
(b)
Parental genotypes: $X^B X^B$ (Black Female) x $X^O Y$ (Orange Male)
Gametes: $X^B$ x $X^O, Y$
Offspring genotypes: $X^B X^O$ , $X^B Y$
Offspring phenotypes: All females Tortoiseshell, All males Black. [4]

10.
(a) 9:3:3:1 [1]
(b)
Expected values:
Grey Long: $9/16 \times 990 = 556.875$
Grey Vestigial: $3/16 \times 990 = 185.625$
Ebony Long: $3/16 \times 990 = 185.625$
Ebony Vestigial: $1/16 \times 990 = 61.875$

Calculations:
Grey Long: $(548-556.875)^2 / 556.875 = 0.14$
Grey Vestigial: $(185-185.625)^2 / 185.625 = 0.002$
Ebony Long: $(192-185.625)^2 / 185.625 = 0.22$
Ebony Vestigial: $(65-61.875)^2 / 61.875 = 0.16$
$\chi^2 = 0.14 + 0.002 + 0.22 + 0.16 \approx 0.52$ [4]
(c) Null Hypothesis: There is no significant difference between the observed and expected results (or the genes assort independently). [1]
(d) Calculated $\chi^2$ (0.52) is less than critical value (7.82). [1] Fail to reject null hypothesis. The difference is due to chance. Genes are likely unlinked. [1] [2]

11.
(a) Base substitution mutation [1] leading to a change in amino acid sequence (Glutamic acid to Valine) [1]. [2]
(b) Heterozygotes ( $Hb^A Hb^S$ ) have resistance to malaria [1]. In malarial regions, they have a survival advantage over $Hb^A Hb^A$ (susceptible to malaria) and $Hb^S Hb^S$ (sickle cell disease) [1]. They survive to reproduce and pass on the $Hb^S$ allele [1]. [3]

12.
(a) A: Histone proteins [1]
B: DNA [1] [2]
(b) Chromatin condenses into chromosomes during mitosis (gene expression stops) [1]. During interphase, chromatin is less condensed (euchromatin), allowing transcription factors and RNA polymerase to access DNA for transcription [1]. [2]

13.
(a) $q^2 = 900/10000 = 0.09$ .
$q = \sqrt{0.09} = 0.3$ . [2]
(b) $p = 1 - q = 0.7$ .
Heterozygotes ( $2pq$ ) = $2 \times 0.7 \times 0.3 = 0.42$ .
Number of carriers = $0.42 \times 10000 = 4200$ . [2]

14.
(a) The proportion of individuals with a specific genotype who express the associated phenotype. [1]
(b) Other genes (modifier genes) may affect limb development [1]. Environmental factors like temperature or nutrient availability during embryonic development may influence expression [1]. [2]

15.
(a) Female $X^r X^r$ x Male $X^R Y$ .
Daughters: $X^R X^r$ (Red eyes). Sons: $X^r Y$ (White eyes).
Phenotypes: All females red-eyed, all males white-eyed. [2]
(b) F1 Female $X^R X^r$ x F1 Male $X^r Y$ .
Offspring:
$X^R X^r$ (Red Female)
$X^r X^r$ (White Female)
$X^R Y$ (Red Male)
$X^r Y$ (White Male)
Ratio: 1 Red Female : 1 White Female : 1 Red Male : 1 White Male. [2]

16.
(a) Enzyme activity increases with temperature up to an optimum (35°C) [1], then decreases rapidly/denatures at higher temperatures [1]. [2]
(b) The enzyme is temperature-sensitive [1]. Extremities (ears, nose, feet) are cooler than the body core [1]. At lower temperatures, the enzyme is active and produces melanin (black fur) [1]. At higher body core temperatures, the enzyme is inactive/denatured, so no melanin is produced (white fur) [1]. [3]

17.
(a) Different mutations may affect the protein structure/function to different extents [1]. Some may prevent protein folding, others may reduce channel conductivity, others may prevent trafficking to the membrane [1]. This leads to varying residual function and symptom severity [1]. [3]
(b) Method: Use of a viral vector (e.g., adenovirus) to deliver the gene via inhalation/aerosol [1].
Ethical concern: Risk of immune response to the vector [1] or insertional mutagenesis (activating oncogenes) [1]. Discussion of long-term safety vs immediate benefit [1]. [4]

18.
(a) Recessive epistasis [1].
(b)
Cross $AaBb \times AaBb$ .
9 $A-B-$ (Purple)
3 $A-bb$ (Red)
3 $aaB-$ (White)
1 $aabb$ (White)
Phenotypes:
Purple: 9
Red: 3
White: $3+1=4$
Ratio: 9 Purple : 3 Red : 4 White. [4]

19.
(a) Mitochondria in the zygote come from the cytoplasm of the egg cell [1]. Sperm mitochondria are generally degraded or do not enter the egg [1]. [2]
(b) Probability is 100% (or 1) for all children [1]. Because the mother passes her mitochondria to all offspring [1]. Mitochondrial diseases are maternally inherited [1]. [3]

20.
Advantages:

Short generation time allows study of many generations quickly [1].
Large number of offspring provides statistically significant data [1].
Easier to control environmental variables (diet, temperature) than in humans [1].
Ethical: Fewer ethical restrictions on breeding and genetic manipulation compared to humans [1].
Many genes are conserved (homologous) between model organisms and humans, allowing insights into human disease [1].

Disadvantages:

Differences in physiology/metabolism may limit direct application to humans [1].
Complex human traits (e.g., behaviour, cognition) may not be fully modelled in simple organisms [1].

Quality of Explanation:

Clear structure comparing advantages and disadvantages.
Specific examples (e.g., Drosophila eye colour vs human genetic disease).
Balanced argument.
[6]
(Marking: 1 mark per valid point, up to 6. Must include at least one advantage and one disadvantage.)