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A Level H1 Biology Genetics Inheritance Quiz

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A-Level Biology H1 Quiz - Genetics Inheritance

Name: __________________________
Class: __________________________
Date: __________________________
Score: _______ / 60

Duration: 60 minutes
Total Marks: 60

Instructions:

Answer all questions.
Write your answers in the spaces provided.
The number of marks is indicated in brackets [ ] at the end of each question or part question.
You are advised to spend approximately 3 minutes per mark.

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

1. Which of the following best describes the term 'allele'?
A. A pair of genes located at the same locus on homologous chromosomes.
B. An alternative form of a gene at a specific locus.
C. The physical expression of a genetic trait.
D. The complete set of genetic material in an organism.
[1]

2. In a test cross, an individual with a dominant phenotype is crossed with an individual that is:
A. Homozygous dominant.
B. Heterozygous.
C. Homozygous recessive.
D. Hemizygous.
[1]

3. A gene has two alleles, $A$ and $a$ . In a population, the frequency of allele $A$ is 0.7. Assuming the population is in Hardy-Weinberg equilibrium, what is the expected frequency of heterozygous individuals ( $Aa$ )?
A. 0.21
B. 0.42
C. 0.49
D. 0.58
[1]

4. Which of the following conditions is necessary for a trait to be considered sex-linked?
A. The gene is located on an autosome.
B. The gene is located on the X or Y chromosome.
C. The trait is only expressed in males.
D. The trait shows codominance.
[1]

5. During meiosis, crossing over occurs during which stage?
A. Prophase I
B. Metaphase I
C. Anaphase II
D. Telophase II
[1]

Section B: Structured Questions – Mendelian and Non-Mendelian Inheritance (Questions 6–10)

6. In pea plants, the allele for tall stems ( $T$ ) is dominant to the allele for short stems ( $t$ ). Two tall plants are crossed, and some of the offspring are short.

(a) State the genotypes of the two parent plants.

[1]

(b) Construct a genetic diagram to show the cross between these two parents. Include parental genotypes, gametes, and offspring genotypes and phenotypes.

[3]

[1]

7. In humans, the ABO blood group system is controlled by three alleles: $I^A$ , $I^B$ , and $I^O$ . $I^A$ and $I^B$ are codominant, and both are dominant to $I^O$ .

(a) Explain what is meant by the term 'codominance'.

[2]

(b) A man with blood group A (heterozygous) marries a woman with blood group B (heterozygous). Determine the possible blood groups of their children. Show your working.

[3]

8. In snapdragons (Antirrhinum), flower color is determined by two alleles, $C^R$ (red) and $C^W$ (white). Heterozygous plants ( $C^R C^W$ ) have pink flowers.

(a) Name this pattern of inheritance.

[1]

(b) A pink-flowered plant is crossed with a white-flowered plant. Predict the phenotypic ratio of the offspring.

[2]

9. Define the term 'linked genes'.

[2]

10. Explain why linked genes do not assort independently during meiosis.

[2]

Section C: Data Interpretation and Pedigree Analysis (Questions 11–15)

11. Fig. 1 shows a pedigree chart for a family affected by a rare genetic condition.

(Note: Imagine a pedigree where I-1 and I-2 are unaffected, but they have an affected daughter II-2. II-2 marries an unaffected male, and they have unaffected children.)

(a) Deduce whether the condition is dominant or recessive. Explain your answer with reference to individuals I-1, I-2, and II-2.

[2]

(b) Deduce whether the condition is autosomal or sex-linked. Explain your answer.

[2]

[1]

12. Haemophilia is a sex-linked recessive condition. Let $X^H$ represent the normal allele and $X^h$ represent the haemophilia allele.

(a) Why are males more likely to suffer from haemophilia than females?

[2]

(b) A carrier female ( $X^H X^h$ ) has children with a normal male ( $X^H Y$ ). Calculate the probability that they will have a son with haemophilia.

[1]

13. In fruit flies (Drosophila), grey body color ( $G$ ) is dominant to black body color ( $g$ ), and long wings ( $L$ ) are dominant to vestigial wings ( $l$ ). These genes are autosomal.

A heterozygous grey-bodied, long-winged fly was test-crossed. The results are shown in Table 1.

Table 1

Phenotype	Number of Offspring
Grey body, long wings	415
Black body, vestigial wings	405
Grey body, vestigial wings	95
Black body, long wings	85

(a) If the genes were unlinked, what would be the expected phenotypic ratio?

[1]

(b) Suggest why the observed numbers differ from the expected ratio for unlinked genes.

[2]

14. Fig. 2 shows the results of a chi-squared ( $\chi^2$ ) test performed on a monohybrid cross. The calculated $\chi^2$ value is 2.5. The critical value at $p=0.05$ with 1 degree of freedom is 3.84.

(a) State the null hypothesis for this test.

[1]

(b) Interpret the result. Is the difference between observed and expected results significant? Explain.

[2]

15. A mutation in a gene results in the substitution of one amino acid in the primary structure of a protein.

(a) Explain how this substitution might alter the tertiary structure of the protein.

[2]

(b) Suggest why some base substitutions do not result in a change to the amino acid sequence.

[1]

Section D: Extended Response and Synthesis (Questions 16–20)

16. Describe the process of meiosis and explain its importance in maintaining genetic variation in a species.

[6]

17. Explain the difference between continuous and discontinuous variation, giving one biological example for each.

[4]

18. In cats, the gene for coat color is located on the X chromosome. The allele for black fur ( $X^B$ ) is codominant with the allele for orange fur ( $X^O$ ). Heterozygous females ( $X^B X^O$ ) have tortoiseshell coats. Males 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. Predict the phenotypes of the offspring. Show your genetic diagram.

[4]

19. Polygenic inheritance involves the interaction of two or more genes to control a single trait.

(a) Describe how polygenic inheritance results in continuous variation.

[2]

(b) Human skin color is an example of polygenic inheritance. Explain how environmental factors can also influence this phenotype.

[2]

20. Epistasis occurs when one gene masks the expression of another gene. In Labrador retrievers, coat color is controlled by two genes:

Gene B: Black ( $B$ ) is dominant to brown ( $b$ ).
Gene E: Pigment deposition ( $E$ ) is dominant to no deposition ( $e$ ). If a dog is homozygous recessive ( $ee$ ), it will be yellow regardless of the B genotype.

(a) State the genotype of a brown Labrador.

[1]

(b) Two black Labradors, both heterozygous for both genes ( $BbEe$ ), are crossed. Calculate the expected phenotypic ratio of the offspring.

[3]

End of Quiz

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Answers

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

1. B
[1]

2. C
[1]

3. B
Calculation: $p = 0.7$ , $q = 0.3$ . Heterozygous frequency = $2pq = 2(0.7)(0.3) = 0.42$ .
[1]

4. B
[1]

5. A
[1]

6.
(a) $Tt$ and $Tt$ (Both must be heterozygous to produce short offspring $tt$ ).
[1]

(b)

Parental Genotypes: $Tt \times Tt$
Gametes: $T, t$ and $T, t$
Offspring Genotypes: $TT, Tt, Tt, tt$
Offspring Phenotypes: 3 Tall : 1 Short
(1 mark for correct gametes, 1 mark for correct offspring genotypes, 1 mark for correct phenotypes/ratio)
[3]

7.
(a) Both alleles are expressed in the phenotype of the heterozygote. Neither allele is recessive to the other.
[2]

(b)

Parental Genotypes: $I^A I^O \times I^B I^O$
Gametes: $(I^A, I^O)$ and $(I^B, I^O)$
Offspring Genotypes: $I^A I^B$ (Group AB), $I^A I^O$ (Group A), $I^B I^O$ (Group B), $I^O I^O$ (Group O)
Possible Blood Groups: A, B, AB, O
(1 mark for correct parental genotypes/gametes, 1 mark for correct offspring genotypes, 1 mark for correct list of blood groups)
[3]

8.
(a) Incomplete dominance
[1]

(b)

Cross: $C^R C^W \times C^W C^W$
Offspring: $C^R C^W$ (Pink) and $C^W C^W$ (White)
Ratio: 1 Pink : 1 White
[2]

9. Genes located on the same chromosome.
[2]

10. Because they are physically connected on the same DNA molecule/chromosome, they tend to be inherited together unless separated by crossing over. They do not align independently at the equator during Metaphase I in the same way non-linked genes on different chromosomes do.
[2]

11.
(a) Recessive. Unaffected parents (I-1 and I-2) have an affected child (II-2). If it were dominant, at least one parent would have to be affected.
[2]

(b) Autosomal. If it were X-linked recessive, the affected daughter (II-2, $X^a X^a$ ) would require her father (I-1) to be affected ( $X^a Y$ ). Since I-1 is unaffected, it cannot be X-linked recessive.
[2]

12.
(a) Males have only one X chromosome (hemizygous). If they inherit the recessive allele ( $X^h$ ), they will express the trait. Females have two X chromosomes and need two recessive alleles ( $X^h X^h$ ) to express the trait, which is statistically less likely.
[2]

(b) 25% or 0.25 or 1/4.
(Cross: $X^H X^h \times X^H Y \rightarrow X^H X^H, X^H X^h, X^H Y, X^h Y$ . Only $X^h Y$ is an affected son. 1 out of 4 total offspring, or 1 out of 2 sons. Question asks for probability of having a son with haemophilia among all children unless specified "given it is a son". Standard interpretation: 1/4 of total births. If interpreted as "among sons", it is 1/2. Given "probability that they will have a son with haemophilia", 1/4 is the standard expectation for total outcome).
[1]

13.
(a) 1:1:1:1
[1]

(b) The genes are linked on the same chromosome. The high numbers of parental types (Grey/Long and Black/Vestigial) indicate that the alleles $G$ and $L$ are on one chromosome and $g$ and $l$ are on the homologous chromosome. The lower numbers are recombinants formed by crossing over.
[2]

(c)

Total offspring = $415 + 405 + 95 + 85 = 1000$
Recombinants = $95 + 85 = 180$
Recombination Frequency = $(180 / 1000) \times 100 = 18\%$
[2]

14.
(a) There is no significant difference between the observed and expected results (any difference is due to chance).
[1]

(b) The calculated value (2.5) is less than the critical value (3.84). Therefore, the null hypothesis is accepted. The difference is not significant.
[2]

15.
(a) The substitution changes the R-group of the amino acid. This may alter the interactions (hydrogen bonds, ionic bonds, disulfide bridges) that maintain the tertiary structure, causing the protein to fold differently.
[2]

(b) The genetic code is degenerate/redundant. More than one triplet code can code for the same amino acid.
[1]

16.

Description: Meiosis involves one round of DNA replication followed by two divisions. Homologous chromosomes pair up in Prophase I. Crossing over occurs between non-sister chromatids, exchanging genetic material. In Metaphase I, homologous pairs align randomly at the equator (independent assortment). In Anaphase I, homologous chromosomes separate. In Meiosis II, sister chromatids separate.
Importance: Crossing over creates new combinations of alleles on chromosomes. Independent assortment creates new combinations of maternal and paternal chromosomes in gametes. Random fertilization further increases variation. This variation allows populations to adapt to changing environments.
(Up to 4 marks for description, 2 marks for explanation of variation)
[6]

17.

Discontinuous Variation: Distinct categories with no intermediates. Controlled by one or a few genes. Little environmental influence. Example: Blood group, gender.
Continuous Variation: A range of phenotypes with no distinct categories. Controlled by many genes (polygenic). Significant environmental influence. Example: Height, mass.
(2 marks for each description with example)
[4]

18.
(a) Males have only one X chromosome ( $XY$ ). They can only carry one allele ( $X^B$ or $X^O$ ). Codominance requires two different alleles to be present in the same individual ( $X^B X^O$ ), which is only possible in females ( $XX$ ).
[2]

(b)

Parental Genotypes: Black Female ( $X^B X^B$ ) $\times$ Orange Male ( $X^O Y$ )
Gametes: Female ( $X^B$ ), Male ( $X^O, Y$ )
Offspring Genotypes: $X^B X^O$ (Female), $X^B Y$ (Male)
Offspring Phenotypes: All females are Tortoiseshell; All males are Black.
(1 mark for parental genotypes, 1 mark for gametes/offspring genotypes, 2 marks for correct phenotypes)
[4]

19.
(a) Multiple genes contribute additively to the phenotype. Each dominant allele adds a small amount to the trait. This results in a bell-shaped distribution of phenotypes rather than distinct classes.
[2]

(b) Exposure to sunlight (UV radiation) stimulates melanin production. Individuals with the same genetic potential for skin color may have darker skin if they are exposed to more sunlight.
[2]

20.
(a) $bbE-$ (specifically $bbEE$ or $bbEe$ ). Must be $bb$ for brown and have at least one $E$ for pigment.
[1]

(b)

Cross: $BbEe \times BbEe$
This is a dihybrid cross with epistasis.
Expected Mendelian Ratio: 9 $B-E-$ : 3 $B-ee$ : 3 $bbE-$ : 1 $bbee$
Phenotypes:
- $B-E-$ (Black): 9
- $B-ee$ (Yellow): 3
- $bbE-$ (Brown): 3
- $bbee$ (Yellow): 1
Combined Yellow: $3 + 1 = 4$
Final Ratio: 9 Black : 3 Brown : 4 Yellow
(1 mark for identifying epistatic effect on yellow, 1 mark for correct grouping, 1 mark for final ratio)
[3]