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A Level H1 Biology Genetics Inheritance Quiz

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Questions

A-Level Biology H1 Quiz - Genetics Inheritance

Name: _________________________
Class: _________________________
Date: _________________________
Score: _______ / 50

Duration: 45 minutes
Total Marks: 50

Instructions:

Answer all questions.
Write your answers in the spaces provided.
The number of marks for each question or part question is given in brackets [ ].
Use appropriate biological terminology and construct genetic diagrams where required.

Section A: Multiple Choice & Short Answer (Questions 1–5)

Answer all questions in this section.

1. Which of the following correctly describes the structure of a nucleotide in DNA?
A. A ribose sugar, a phosphate group, and a nitrogenous base.
B. A deoxyribose sugar, a phosphate group, and a nitrogenous base.
C. A deoxyribose sugar, two phosphate groups, and a nitrogenous base.
D. A ribose sugar, a phosphate group, and two nitrogenous bases.
[1]

2. In a double-stranded DNA molecule, if 20% of the bases are adenine, what percentage of the bases are cytosine?
A. 20%
B. 30%
C. 40%
D. 60%
[1]

3. During DNA replication, which enzyme is responsible for joining Okazaki fragments on the lagging strand?
A. DNA helicase
B. DNA polymerase
C. DNA ligase
D. Primase
[1]

4. A mutation results in the substitution of a single nucleotide base, but the amino acid sequence of the resulting protein remains unchanged. This is known as a:
A. Missense mutation
B. Nonsense mutation
C. Silent mutation
D. Frameshift mutation
[1]

5. State the specific type of bond that holds the two complementary strands of DNA together.

[1]

Section B: Structured Questions (Questions 6–12)

Answer all questions in this section.

6. The table below shows the results of an experiment investigating the effect of temperature on the activity of an enzyme involved in DNA replication.

Temperature (°C)	Relative Enzyme Activity (%)
20	30
30	65
37	100
45	80
60	10
80	0

(a) Explain why the enzyme activity decreases significantly at 80°C.

[2]

(b) Suggest why the enzyme activity at 20°C is lower than at 37°C, referring to molecular movement.

[2]

7. Figure 1 represents a simplified diagram of transcription.

(Imagine Fig 1 showing a DNA template strand, an mRNA strand being synthesized, and an enzyme labeled 'X')

(a) Name enzyme X.

[1]

(b) State the direction in which enzyme X moves along the DNA template strand.

[1]

[2]

8. In humans, the ability to roll the tongue is controlled by a single gene with two alleles. The allele for tongue rolling (R) is dominant to the allele for non-rolling (r).

(a) Two parents who can both roll their tongues have a child who cannot roll their tongue.
(i) State the genotypes of the parents.
Parent 1: _______________ Parent 2: _______________
[1]

(ii) Construct a genetic diagram to show how these parents can produce a non-rolling child. Include parental genotypes, gametes, and offspring genotypes/phenotypes.
[3]

(b) The couple expects another child. Calculate the probability that this child will be able to roll their tongue.
Probability: _______________
[1]

9. Haemophilia is a sex-linked recessive condition caused by an allele on the X chromosome. Let $X^H$ represent the normal allele and $X^h$ represent the haemophilia allele.

(a) Explain why haemophilia is more common in males than in females.

[2]

(b) A carrier female ( $X^H X^h$ ) marries a normal male ( $X^H Y$ ).
Construct a genetic diagram to determine the possible genotypes and phenotypes of their offspring.
[4]

10. Describe the process of translation, focusing on the role of tRNA and ribosomes.

[4]

11. A gene consists of 1200 nucleotide bases in its coding sequence.

(a) Calculate the maximum number of amino acids that could be present in the polypeptide chain produced from this gene.
Answer: _______________
[1]

(b) Explain why the actual number of amino acids in the functional protein might be lower than this calculated maximum.

[2]

12. Distinguish between the terms gene and allele.

[2]

Section C: Data Interpretation & Extended Response (Questions 13–20)

Answer all questions in this section.

13. Figure 2 shows a pedigree chart for a family affected by a rare genetic disorder.

(Imagine Fig 2: A pedigree where unaffected parents have an affected daughter, and the trait appears in subsequent generations only when both parents are carriers or affected.)

(a) Deduce, with reasons, whether the disorder is dominant or recessive.

[2]

(b) Deduce, with reasons, whether the disorder is autosomal or sex-linked.

[2]

(c) Individual 5 in the pedigree is unaffected but has an affected sibling. State the probability that Individual 5 is a carrier.
Probability: _______________
[1]

14. In a species of flower, petal color is controlled by two alleles, $C^R$ (red) and $C^W$ (white). These alleles show codominance.

(a) State the phenotype of a plant with the genotype $C^R C^W$ .

[1]

(b) A red-flowered plant is crossed with a white-flowered plant. All F1 offspring have pink flowers.
Construct a genetic diagram to show the expected phenotypic ratio when two F1 plants are crossed.
[4]

15. Explain the significance of the semi-conservative nature of DNA replication.

[3]

16. A mutation in the CFTR gene causes Cystic Fibrosis. This mutation involves the deletion of three nucleotide bases, resulting in the loss of a single amino acid (phenylalanine) from the protein.

(a) Why does the deletion of three bases not cause a frameshift mutation?

[2]

(b) Suggest how the loss of a single amino acid might affect the function of the CFTR protein.

[2]

17. Describe the role of mRNA in protein synthesis.

[3]

18. In cats, coat color is determined by a gene on the X chromosome. The allele $X^B$ codes for black fur, and $X^O$ codes for orange fur. Female cats heterozygous for these alleles ( $X^B X^O$ ) have tortoiseshell coats (patches of black and orange).

(a) Explain why male cats cannot have a tortoiseshell coat (assuming normal chromosomal composition).

[2]

(b) A black female cat is crossed with an orange male cat.
Construct a genetic diagram to show the expected phenotypes of the offspring.
[4]

19. Compare and contrast the structures of DNA and RNA.

[4]

20. "Mutations are always harmful to an organism."
Discuss this statement, providing examples to support your answer.

[5]

END OF QUIZ

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Answers

A-Level Biology H1 Quiz - Genetics Inheritance (Answer Key)

1. B
[1]

2. B
Reasoning: If A = 20%, then T = 20%. Total A+T = 40%. Remaining 60% is G+C. Since G=C, C = 30%.
[1]

3. C
[1]

4. C
[1]

5. Hydrogen bonds
[1]

6.
(a) At 80°C, the high kinetic energy breaks the hydrogen bonds and other bonds maintaining the tertiary structure of the enzyme [1]. The active site changes shape (denaturation), so the substrate can no longer bind [1].
(b) At 20°C, molecules have less kinetic energy [1]. There are fewer successful collisions between enzyme and substrate per unit time compared to 37°C [1].
[2] + [2]

7.
(a) RNA polymerase
[1]
(b) 3' to 5' (along the template strand)
[1]
(c) DNA is located in the nucleus and cannot leave [1]. mRNA carries the genetic code from the nucleus to the ribosomes in the cytoplasm for translation [1].
[2]

8.
(a) (i) Parent 1: Rr, Parent 2: Rr
[1]
(ii)
Parental Phenotype: Roller x Roller
Parental Genotype: Rr x Rr
Gametes: R, r x R, r
Offspring Genotypes: RR, Rr, Rr, rr
Offspring Phenotypes: Roller, Roller, Roller, Non-roller
Ratio: 3 Roller : 1 Non-roller
[Award marks for correct gametes, correct offspring genotypes, and correct phenotypes.]
[3]
(b) 75% or 0.75 or 3/4
[1]

9.
(a) Males have only one X chromosome (XY) [1]. If they inherit the recessive allele ( $X^h$ ), they will express the trait because there is no corresponding allele on the Y chromosome to mask it [1]. Females have two X chromosomes and need two recessive alleles to express the trait.
[2]
(b)
Parental Genotype: $X^H X^h$ x $X^H Y$
Gametes: $X^H$ , $X^h$ x $X^H$ , $Y$
Offspring Genotypes:
$X^H X^H$ (Normal Female)
$X^H X^h$ (Carrier Female)
$X^H Y$ (Normal Male)
$X^h Y$ (Haemophiliac Male)
Phenotypes: 1 Normal Female : 1 Carrier Female : 1 Normal Male : 1 Haemophiliac Male
[4]

10.
mRNA binds to the ribosome [1]. tRNA molecules with specific anticodons bring specific amino acids to the ribosome [1]. The anticodon of the tRNA pairs with the complementary codon on the mRNA [1]. Peptide bonds form between adjacent amino acids, forming a polypeptide chain [1].
[4]

11.
(a) 400
Reasoning: 1200 bases / 3 bases per codon = 400 amino acids.
[1]
(b) The gene contains introns (non-coding regions) that are removed during post-transcriptional modification [1]. Also, the stop codon does not code for an amino acid [1].
[2]

12.
A gene is a sequence of DNA bases that codes for a polypeptide/protein [1]. An allele is an alternative form of a gene [1].
[2]

13.
(a) Recessive [1]. Unaffected parents (e.g., individuals 1 and 2) have an affected child (individual 3), which is only possible if both parents are carriers of a recessive allele [1].
[2]
(b) Autosomal [1]. If it were X-linked recessive, the affected daughter (individual 3) would have an affected father (individual 1), but the father is unaffected [1].
[2]
(c) 2/3
Reasoning: Parents are heterozygous (Aa). Unaffected offspring can be AA or Aa. Ratio of AA:Aa is 1:2. Probability of being Aa is 2/3.
[1]

14.
(a) Pink
[1]
(b)
Parental Phenotype: Pink x Pink
Parental Genotype: $C^R C^W$ x $C^R C^W$
Gametes: $C^R$ , $C^W$ x $C^R$ , $C^W$
Offspring Genotypes: $C^R C^R$ , $C^R C^W$ , $C^R C^W$ , $C^W C^W$
Offspring Phenotypes: Red, Pink, Pink, White
Phenotypic Ratio: 1 Red : 2 Pink : 1 White
[4]

15.
Semi-conservative replication means each new DNA molecule contains one original (parental) strand and one newly synthesized strand [1]. This ensures that the genetic information is preserved accurately [1]. It allows for proofreading and repair mechanisms to use the original strand as a template [1].
[3]

16.
(a) Because the genetic code is read in triplets (codons) [1]. Deleting three bases removes exactly one codon, so the reading frame for the subsequent bases remains unchanged [1].
[2]
(b) The loss of an amino acid may change the primary structure [1]. This can alter the tertiary structure (folding) of the protein, potentially affecting the shape of the active site or channel, leading to loss of function [1].
[2]

17.
mRNA carries the genetic code from DNA in the nucleus to the ribosomes in the cytoplasm [1]. It contains codons (triplets of bases) that specify the sequence of amino acids [1]. It serves as the template for translation [1].
[3]

18.
(a) Males have only one X chromosome (XY) [1]. They can only have one allele for coat color ( $X^B$ or $X^O$ ), so they cannot be heterozygous and express both colors simultaneously [1].
[2]
(b)
Parental Phenotype: Black Female x Orange Male
Parental Genotype: $X^B X^B$ x $X^O Y$
Gametes: $X^B$ x $X^O$ , $Y$
Offspring Genotypes: $X^B X^O$ , $X^B Y$
Offspring Phenotypes: Tortoiseshell Female, Black Male
[4]

19.
Similarities: Both are nucleic acids; both contain phosphate groups, pentose sugars, and nitrogenous bases (A, G, C) [1].
Differences:

DNA contains deoxyribose sugar; RNA contains ribose sugar [1].
DNA contains thymine (T); RNA contains uracil (U) [1].
DNA is double-stranded (helix); RNA is usually single-stranded [1].
[4]

20.
Argument against:

Some mutations are silent (no change in amino acid sequence) and have no effect [1].
Some mutations can be beneficial, providing an advantage in specific environments (e.g., antibiotic resistance in bacteria, sickle cell trait providing malaria resistance) [1].
Mutations are the source of genetic variation, which is essential for evolution by natural selection [1].

Argument for:

Many mutations disrupt protein function, leading to genetic disorders or cell death (e.g., Cystic Fibrosis, Cancer) [1].
Mutations in regulatory genes can lead to uncontrolled cell division [1].

Conclusion:
The statement is incorrect. While many mutations are neutral or harmful, some are beneficial and essential for the survival of species in changing environments [1].
[5]