A Level H2 Biology Genetics Inheritance Quiz

A-Level Biology H2 Quiz - Genetics Inheritance

Answer Key and Marking Scheme

Section A: Multiple Choice Questions (10 marks)

1. A [1]
In a dihybrid cross (AaBb × AaBb), the probability of being homozygous recessive for both traits (aabb) is 1/4 × 1/4 = 1/16. Each gene segregates independently, so the probability of aa is 1/4 and bb is 1/4; multiply for both together.

2. B [1]
A cross between Aa and aa produces offspring with genotypes Aa and aa in a 1:1 ratio. The probability of one offspring showing the dominant phenotype (Aa) is 1/2. For two offspring both to show the dominant phenotype: 1/2 × 1/2 = 1/4.

3. B [1]
Codominance occurs when both alleles in a heterozygote are fully and equally expressed, resulting in a phenotype that shows both traits simultaneously (e.g., AB blood group where both A and B antigens are expressed). Option A describes complete dominance; Option C describes incomplete dominance; Option D describes pleiotropy.

4. A [1]
The man has genotype $I^A I^B$ and the woman has genotype $ii$. The possible gametes are $I^A$ and $I^B$ from the father, and $i$ from the mother. Offspring genotypes: $I^A i$ (blood group A) and $I^B i$ (blood group B), in equal proportions. Blood groups AB and O are not possible.

5. B [1]
A 9:3:3:1 phenotypic ratio in a dihybrid cross is the classic result of two independently assorting genes, each showing complete dominance. This was first described by Gregor Mendel. Linked genes would produce a deviation from this ratio; epistasis would modify the ratio; incomplete dominance would produce different ratios.

6. B [1]
Recombination frequency = (number of recombinant offspring / total offspring) × 100. Recombinants are Grey body, vestigial wings (10) and Black body, long wings (10). Total = 100. RF = (10 + 10) / 100 × 100 = 20%.

7. C [1]
A carrier female ($X^H X^h$) produces 50% $X^H$ and 50% $X^h$ gametes. Sons inherit the X chromosome from the mother, so there is a 50% chance a son inherits $X^h$ and has haemophilia. Option A is incorrect because X-linked recessive disorders are more common in males. Option B is incorrect because fathers pass Y to sons, not X. Option D is incorrect because mothers do not have Y chromosomes.

8. C [1]
Given q = 0.3, then p = 1 − 0.3 = 0.7. Frequency of heterozygotes = 2pq = 2 × 0.7 × 0.3 = 0.42.

9. C [1]
The original codon TGG codes for tryptophan. The mutated codon TAG is a stop codon (amber). This is a nonsense mutation because it introduces a premature stop codon, resulting in a truncated protein. It is not a frameshift because only one nucleotide is substituted (not inserted or deleted).

10. C [1]
Human skin colour is controlled by multiple genes (estimated 3–6 genes), each contributing additively to the phenotype. This is a classic example of polygenic inheritance, resulting in a continuous range of phenotypes. ABO blood groups involve multiple alleles of a single gene; snapdragon flower colour shows incomplete dominance; sickle cell anaemia is a single-gene disorder.

Section B: Structured Questions (30 marks)

11.

(a) Epistasis [1]
Epistasis occurs when the expression of one gene affects (masks or modifies) the expression of another gene. Here, gene B can only produce red pigment if gene A has already produced the pink intermediate.

(b) [3]

Completed Punnett square:

Phenotypic ratio: 9 Red : 3 Pink : 4 White [1 for correct ratio]

• Red (A_B_): 9/16 — both functional enzymes present
• Pink (A_bb): 3/16 — enzyme A functional, enzyme B non-functional
• White (aaB_ + aabb): 4/16 — enzyme A non-functional (no intermediate produced, so B cannot act)

[1 mark for correct Punnett square, 1 mark for correct phenotypic ratio, 1 mark for correct explanation of classes]

(c) [2]
The 9:3:3:1 ratio assumes independent gene interaction where each gene contributes independently to the phenotype. In this case, the genes interact in a biochemical pathway (epistasis). The presence of the recessive homozygous genotype (aa) masks the expression of gene B because no intermediate is produced for enzyme B to act upon. This modifies the expected ratio to 9:3:4 (recessive epistasis).

12.

(a) [2]
Males are hemizygous for X-linked genes (they have only one X chromosome). Therefore, a single recessive allele ($X^h$) on the X chromosome will express the disorder. Females have two X chromosomes and must inherit two copies of the recessive allele ($X^h X^h$) to be affected. A female with one copy ($X^H X^h$) is a carrier and phenotypically normal. This makes the disorder much more common in males.

(b) [4]

Genetic diagram:

(i) Probability that first child will have haemophilia = 1/4 [1]
(Only $X^h Y$ genotype out of 4 possible outcomes)

(ii) Probability that first daughter will be a carrier = 1/2 [1]
(Daughters are $X^H X^H$ or $X^H X^h$; 1 out of 2 daughters is a carrier)

(c) [2]
The probability remains 1/4 [1]. Each fertilisation event is independent — the outcome of one pregnancy does not affect the next. The mother always produces 50% $X^H$ and 50% $X^h$ gametes, and the father always produces 50% $X^H$ and 50% Y gametes. The previous son's genotype does not alter the probabilities for subsequent offspring [1].

13.

(a) [2]
The condition is autosomal [1]. If it were X-linked recessive, the affected male in Generation II would pass the allele to all his daughters (who would be carriers or affected). His daughter in Generation III is affected, which is consistent with autosomal inheritance. If it were X-linked dominant, the affected male would pass it to all daughters — which is observed — but the key evidence against X-linked is that an affected female in Generation III has an unaffected father (the affected male in Gen II married an unaffected female; their affected daughter could inherit the allele from her mother if autosomal recessive). The pattern is most consistent with autosomal recessive: unaffected parents (Gen I) produced an affected child (Gen II-1), indicating both parents were carriers.

(b) [2]
The condition is recessive [1]. The unaffected parents in Generation I produced an affected child (II-1). This is only possible if both parents were heterozygous carriers and the condition is recessive. If the condition were dominant, at least one parent would have to be affected to pass on the dominant allele [1].

(c) [2]
Individual II-2 is unaffected. Since her brother (II-1) is affected (aa), and her parents are both carriers (Aa), the possible genotypes for II-2 are: 1/3 AA or 2/3 Aa [1]. She marries an unaffected male from the general population. Assuming the condition is rare, the male is most likely AA. If II-2 is Aa (probability 2/3), the cross Aa × AA produces 0 affected offspring. If II-2 is AA (probability 1/3), all offspring are unaffected. Therefore, the probability that their first child will be affected is 0 (or effectively 0, assuming the population male is not a carrier) [1].

14.

(a) [3]
Total alleles in population = 500 × 2 = 1000

Number of M alleles = (180 × 2) + 240 = 360 + 240 = 600
Frequency of M (p) = 600/1000 = 0.6 [1]

Number of m alleles = (80 × 2) + 240 = 160 + 240 = 400
Frequency of m (q) = 400/1000 = 0.4 [1]

Check: p + q = 0.6 + 0.4 = 1.0 ✓ [1]

(b) [2]
Expected frequency of heterozygotes (2pq) = 2 × 0.6 × 0.4 = 0.48 [1]
Expected number of heterozygotes = 0.48 × 500 = 240 [1]

(c) [2]
Any two of the following (1 mark each, max 2):

• No mutation
• No natural selection (all genotypes have equal fitness)
• Random mating (no sexual selection)
• Large population size (no genetic drift)
• No gene flow (no migration into or out of the population)

15.

(a) Gene mutation / point mutation / substitution mutation [1]

(b) [3]
The substitution of valine (hydrophobic/non-polar) for glutamic acid (hydrophilic/negatively charged) at position 6 of the β-globin polypeptide [1] causes a hydrophobic patch to appear on the surface of the haemoglobin molecule [1]. Under low oxygen conditions, these hydrophobic patches cause haemoglobin molecules to aggregate and polymerise, forming long fibres that distort the red blood cell into a sickle shape [1].

(c) [2]
Heterozygous individuals have a selective advantage in malaria-endemic regions because the sickle cell trait (Hb^A Hb^S) provides partial resistance to Plasmodium falciparum infection [1]. This is an example of heterozygote advantage (balancing selection), where the Hb^S allele is maintained in the population despite being harmful in the homozygous state, because heterozygotes have higher fitness than either homozygote in malaria-affected areas [1].

Section C: Data Interpretation and Extended Response (20 marks)

16.

(a) [2]
In Cross 1, when pure-breeding round yellow is crossed with wrinkled green, all F1 offspring are round yellow [1]. This indicates that round is dominant over wrinkled, and yellow is dominant over green, because the recessive traits (wrinkled and green) are masked in the heterozygous F1 generation [1].

(b) [2]
The deviation from the expected 1:1:1:1 ratio indicates that the two genes are linked (located on the same chromosome) [1]. The parental phenotypes (round yellow and wrinkled green) appear in much higher numbers than the recombinant phenotypes (round green and wrinkled yellow), showing that the alleles tend to be inherited together rather than assorting independently [1].

(c) [2]
Recombination frequency = (recombinants / total) × 100
Recombinants = Round green (5) + Wrinkled yellow (3) = 8
Total = 100
RF = (8/100) × 100 = 8% [2]

(d) [2]
A recombination frequency of 8% indicates that the two genes are located on the same chromosome and are 8 map units (centimorgans) apart [1]. The low recombination frequency means the genes are closely linked and tend to be inherited together, which explains why the parental combinations are much more frequent than recombinant combinations [1].

17.

(a) [2]
Population A shows a gradual, slight increase in the frequency of the disorder over 10 generations, rising from approximately 0.04 to 0.06 [1]. Population B shows a dramatic increase in the frequency of the disorder, rising from approximately 0.04 to 0.25 over the same period [1].

(b) [2]
Any two of the following (1 mark each):

• Population B may have a smaller effective population size, making it more susceptible to genetic drift
• Population B may have experienced a founder effect, where the founding individuals happened to carry the recessive allele at higher frequency
• Population B may have experienced a population bottleneck, reducing genetic diversity and increasing allele frequency by chance
• Population B may have higher rates of consanguineous mating (inbreeding), increasing homozygosity
• Population B may have different selection pressures (e.g., heterozygote advantage)

(c) [2]
Genetic drift is the random change in allele frequencies due to chance events in small populations [1]. In Population B, the recessive allele frequency may have increased randomly over generations because the population is small, and chance events (such as certain individuals contributing more offspring to the next generation) caused the allele to become more common. This is especially pronounced in isolated populations with limited gene flow [1].

18.

(a) [3]
The ΔF508 mutation is a deletion of three nucleotides (CTT) from the CFTR gene, resulting in the loss of phenylalanine at position 508 of the protein [1]. This deletion occurs in the first nucleotide-binding domain (NBD1) of the CFTR protein, which is essential for ATP binding and channel gating [1]. The misfolded protein is recognised by the cell's quality control system (endoplasmic reticulum-associated degradation) and is degraded before it can reach the cell membrane, resulting in a lack of functional chloride channels at the cell surface [1].

(b) [3]
Given: 400 affected individuals out of 10,000
Frequency of affected (q²) = 400/10,000 = 0.04 [1]
q = √0.04 = 0.2
p = 1 − 0.2 = 0.8
Frequency of carriers (2pq) = 2 × 0.8 × 0.2 = 0.32 [1]
Number of carriers = 0.32 × 10,000 = 3,200 [1]

(c) [2]
Each conception is an independent event [1]. The parental genotypes do not change: both parents are carriers (Cc × Cc). The probability of each offspring inheriting two recessive alleles (cc) is always 1/4, regardless of the genotypes of previous children. The outcome of one pregnancy does not influence the outcome of the next [1].

19.

(a) Recessive epistasis [1]

(b) [4]
Let gene A control the production of grey pigment and gene B control the conversion of grey to black.

• Parental cross: AAbb (black) × aaBB (white) [1]
• F1 generation: AaBb (grey) — both genes functional, grey pigment produced and partially converted [1]
• F2 generation: A_B_ (9/16 grey), A_bb (3/16 black), aaB_ (3/16 white), aabb (1/16 white) [1]

The aa genotype is epistatic — when homozygous recessive (aa), it masks the expression of gene B, resulting in white fur regardless of the B allele. This gives the 9:3:4 ratio (9 grey : 3 black : 4 white) [1].

(c) [2]
F1 (AaBb) × white (aabb):

Offspring: 1 AaBb (grey) : 1 Aabb (black) : 1 aaBb (white) : 1 aabb (white)
Phenotypic ratio: 1 Grey : 1 Black : 2 White [2]

20.

(a) [1]
A single nucleotide polymorphism (SNP) is a variation in a single nucleotide (A, T, C, or G) at a specific position in the genome, that occurs in at least 1% of the population.

(b) [2]
GWAS are useful because they scan the entire genome for SNPs associated with diseases, allowing researchers to identify multiple genetic variants that contribute to complex diseases [1]. However, GWAS may miss rare variants (which are not well-tagged by common SNPs), gene-gene interactions (epistasis), gene-environment interactions, and structural variants (deletions, duplications) that are not captured by SNP arrays [1].

(c) [2]
Complex diseases like diabetes are polygenic, meaning they are influenced by multiple genes, each contributing a small effect to the overall phenotype [1]. Additionally, environmental factors (diet, exercise, lifestyle) interact with genetic predisposition. This multifactorial nature means the phenotype does not segregate into discrete categories as expected under simple Mendelian inheritance, and instead shows a continuous distribution of risk [1].

(d) [1]
Understanding SNPs allows for pharmacogenomics — tailoring drug treatments to an individual's genetic profile. For example, certain SNPs affect how individuals metabolise specific drugs, allowing doctors to prescribe the most effective drug and dosage for each patient, minimising adverse reactions and maximising therapeutic benefit.

END OF ANSWER KEY