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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: ______ / 50

Duration: 45 minutes
Total Marks: 50
Instructions: Answer ALL questions in the spaces provided. Show all working for calculation questions. Marks are indicated in brackets. Use appropriate genetic symbols and terminology throughout.

Section A: Mendelian Genetics and Inheritance Patterns (15 marks)

1. In pea plants, tall stem (T) is dominant over dwarf stem (t), and round seed (R) is dominant over wrinkled seed (r). A plant heterozygous for both traits is test-crossed.

(a) State the genotype of the heterozygous plant and the genotype of the test-cross parent. [2]

(b) Using a Punnett square or the forked-line method, determine the expected phenotypic ratio of the offspring. [3]

2. A couple, both with normal vision, have a son who is colour-blind. Colour-blindness is an X-linked recessive trait.

(a) Using the symbols X^C (normal allele) and X^c (colour-blind allele), state the genotypes of both parents. [2]

(b) Explain why the son is colour-blind even though both parents have normal vision. [2]

3. In a species of flowering plant, flower colour is controlled by a single gene with two alleles. The allele for red flowers (C^R) is codominant with the allele for white flowers (C^W). Heterozygous plants have pink flowers.

(a) Name the type of inheritance pattern described. [1]

(b) A pink-flowered plant is crossed with a white-flowered plant. Predict the phenotypic ratio of the offspring. [3]

4. In pea plants, the allele for purple flowers (P) is dominant over the allele for white flowers (p). A true-breeding purple-flowered plant is crossed with a true-breeding white-flowered plant.

(a) State the genotypes of the parental plants. [1]

(b) Determine the genotype and phenotype of the F1 generation. [1]

5. In humans, free earlobes (F) are dominant over attached earlobes (f). A man with free earlobes, whose mother had attached earlobes, marries a woman with attached earlobes.

(a) State the genotype of the man. Explain your reasoning. [2]

(b) Determine the probability that their first child will have attached earlobes. Show your working. [2]

Section B: Gene Interactions and Modified Ratios (15 marks)

6. In sweet peas, purple flower colour requires at least one dominant allele of both gene A and gene B. Gene A encodes enzyme A, which converts a colourless precursor to a pink intermediate. Gene B encodes enzyme B, which converts the pink intermediate to purple pigment. Homozygous recessive aa or bb results in white flowers.

(a) State the type of gene interaction shown. [1]

(b) Two pure-breeding white-flowered plants (AAbb and aaBB) are crossed. State the genotype and phenotype of the F1 generation. [2]

(c) The F1 plants are self-pollinated. Using a Punnett square, determine the expected phenotypic ratio of the F2 generation. [4]

7. In Labrador retrievers, coat colour is controlled by two genes. Gene E determines whether pigment is deposited in the fur (E = pigment deposited, e = no pigment → yellow coat). Gene B determines the colour of pigment deposited (B = black, b = brown). A dog with genotype ee is yellow regardless of the B gene alleles.

(a) Name the type of epistasis shown. [1]

(b) A black Labrador (EeBb) is crossed with a yellow Labrador (eebb). Predict the phenotypic ratio of the offspring. Show your working. [4]

8. In a certain plant, two pure-breeding white-flowered strains were crossed. All F1 offspring had purple flowers. When the F1 plants were self-pollinated, the F2 generation consisted of 112 purple-flowered plants and 88 white-flowered plants.

(a) State the expected phenotypic ratio if the trait is controlled by two genes with complementary gene action. [1]

(b) Using the chi-squared (χ²) test, determine whether the observed results fit the expected ratio. The chi-squared formula is: χ² = Σ[(O − E)² / E]. Show all your working. [4]

9. In summer squash, fruit colour is controlled by two genes. Gene W determines whether pigment is produced (W = white, w = coloured). Gene Y determines the colour of pigment (Y = yellow, y = green). The dominant W allele is epistatic to gene Y, resulting in white fruit regardless of the Y genotype.

(a) Name the type of epistasis shown. [1]

(b) A white squash plant (WwYy) is crossed with a green squash plant (wwyy). Predict the phenotypic ratio of the offspring. [3]

10. In mice, coat colour is controlled by two genes. Gene C determines pigment production (C = pigment, c = albino). Gene A determines agouti pattern (A = agouti, a = black). The cc genotype is epistatic, resulting in albino mice regardless of the A gene alleles.

(a) State the expected phenotypic ratio in the F2 generation from a dihybrid cross (CcAa × CcAa). [1]

(b) Explain why the ratio differs from the typical 9:3:3:1 Mendelian ratio. [2]

Section C: Chromosomal Inheritance and Genetic Analysis (20 marks)

11. Figure 1 shows a pedigree for a family where some individuals have Huntington's disease, an autosomal dominant disorder.

     I       □───■
            │
     II   □───○   ■───○
         │       │
     III □  ■   ○  □

Key: □ = unaffected male, ○ = unaffected female, ■ = affected male, ● = affected female

(a) Using the symbols H (disease allele) and h (normal allele), deduce the genotype of individual I-2. Explain your reasoning. [2]

(b) Individual III-1 is unaffected. Explain why this is possible even though his father (II-3) is affected. [2]

(c) If individual III-4 marries an unaffected man, calculate the probability that their first child will have Huntington's disease. [2]

12. A researcher studying a rare genetic disorder analysed DNA samples from a family using gel electrophoresis. The disorder is caused by a mutation that creates an additional restriction enzyme recognition site in the gene, resulting in smaller DNA fragments. Figure 2 shows the results.

Individual	Fragment sizes (kb)
Father	8.0, 5.0, 3.0
Mother	8.0, 8.0
Child 1	8.0, 5.0, 3.0
Child 2	8.0, 8.0
Child 3	8.0, 5.0, 3.0

(a) Explain why the father shows three fragments while the mother shows only one fragment. [3]

(b) Deduce the genotype of Child 2 with respect to the disorder. Explain your answer. [2]

(c) The normal allele produces an 8.0 kb fragment. The disease allele produces 5.0 kb and 3.0 kb fragments. State whether the disorder is dominant or recessive. Justify your answer with reference to the data. [2]

13. In Drosophila melanogaster, eye colour is an X-linked trait. Red eyes (X^R) are dominant over white eyes (X^r). A red-eyed female (homozygous) is crossed with a white-eyed male.

(a) State the genotypes of the parental flies. [2]

(b) Predict the phenotypes and their ratios in the F1 generation. [3]

14. In humans, Duchenne muscular dystrophy (DMD) is an X-linked recessive disorder. A woman whose brother has DMD is concerned about passing the disorder to her children. She and her husband are both unaffected.

(a) Using appropriate symbols, determine the possible genotypes of the woman, her husband, and their potential offspring. [4]

(b) Calculate the probability that the couple will have an affected son. [1]

15. In a species of bird, feather colour is a Z-linked trait. The allele for green feathers (Z^G) is dominant over the allele for blue feathers (Z^g). A green-feathered male (heterozygous) is crossed with a blue-feathered female.

(a) State the genotypes of the parental birds. [2]

(b) Predict the phenotypic ratio of the offspring. [3]

Section D: Genetic Variation and Advanced Topics (10 marks)

16. In a population of butterflies, wing colour is controlled by a single gene with two alleles: B (black) and b (white). The population consists of 360 black butterflies and 40 white butterflies. The population is in Hardy-Weinberg equilibrium.

(a) Calculate the frequency of the recessive allele (b). [2]

(b) Calculate the number of heterozygous black butterflies in the population. [3]

17. Explain how crossing over during meiosis contributes to genetic variation in gametes. [3]

18. Distinguish between linked genes and unlinked genes in terms of their inheritance patterns. [3]

19. A man with blood type AB marries a woman with blood type O. State the possible blood types of their children and explain why no other blood types are possible. [2]

20. In a species of plant, leaf shape is controlled by a single gene with two alleles: L (lobed) and l (entire). A population of 500 plants was surveyed, and 80 plants had entire leaves. Assuming Hardy-Weinberg equilibrium, calculate the expected number of heterozygous plants. [2]

END OF QUIZ

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Answers

A-Level Biology H2 Quiz - Genetics Inheritance: Answer Key

Total Marks: 50

Section A: Mendelian Genetics and Inheritance Patterns (15 marks)

1. (a) Heterozygous plant: TtRr; Test-cross parent: ttrr [2 marks – 1 for each correct genotype]

(b) Expected phenotypic ratio: 1 tall round : 1 tall wrinkled : 1 dwarf round : 1 dwarf wrinkled (1:1:1:1) [3 marks – 1 for correct gametes from TtRr (TR, Tr, tR, tr), 1 for correct gamete from ttrr (tr), 1 for correct ratio]

2. (a) Mother: X^C X^c; Father: X^C Y [2 marks – 1 for each correct genotype]

(b) The mother is a carrier (heterozygous X^C X^c). She passed her X^c allele to her son, who received a Y chromosome from his father. Since males have only one X chromosome, the presence of the recessive allele X^c results in colour-blindness. [2 marks – 1 for identifying mother as carrier, 1 for explaining hemizygous expression in males]

(c) Probability = 0 (0%). A daughter receives one X chromosome from each parent. The father (X^C Y) always passes X^C to daughters. Even if the mother passes X^c, the daughter will be heterozygous X^C X^c (carrier, not colour-blind). [2 marks – 1 for correct probability, 1 for explanation]

3. (a) Codominance [1 mark]

(b) Pink-flowered plant: C^R C^W; White-flowered plant: C^W C^W. Gametes from pink: C^R, C^W; Gametes from white: C^W. Offspring: 1 C^R C^W (pink) : 1 C^W C^W (white). Phenotypic ratio: 1 pink : 1 white. [3 marks – 1 for correct parental genotypes, 1 for correct gametes, 1 for correct ratio]

4. (a) Purple-flowered parent: PP; White-flowered parent: pp [1 mark]

(b) F1 genotype: Pp; F1 phenotype: Purple [1 mark]

(c) F1 self-cross: Pp × Pp. Gametes: P, p from each parent. F2 genotypic ratio: 1 PP : 2 Pp : 1 pp. F2 phenotypic ratio: 3 purple : 1 white. [3 marks – 1 for correct gametes, 1 for correct genotypic ratio, 1 for correct phenotypic ratio]

5. (a) The man's genotype is Ff. He has free earlobes (dominant phenotype) so he must have at least one F allele. His mother had attached earlobes (ff), so she could only pass on an f allele. Therefore, the man must have inherited an f allele from his mother, making him heterozygous Ff. [2 marks – 1 for correct genotype, 1 for reasoning]

(b) Man (Ff) × Woman (ff). Gametes from man: F, f. Gametes from woman: f. Offspring: 1/2 Ff (free earlobes), 1/2 ff (attached earlobes). Probability of attached earlobes = 1/2 (50%). [2 marks – 1 for correct cross, 1 for correct probability]

Section B: Gene Interactions and Modified Ratios (15 marks)

6. (a) Complementary gene action (or duplicate recessive epistasis) [1 mark]

(b) F1 genotype: AaBb; F1 phenotype: Purple [2 marks – 1 for genotype, 1 for phenotype]

A_B_ (purple): 9/16
A_bb (white): 3/16
aaB_ (white): 3/16
aabb (white): 1/16 Phenotypic ratio: 9 purple : 7 white. [4 marks – 1 for correct gametes, 1 for correct Punnett square/setup, 1 for identifying purple genotypes, 1 for correct ratio]

7. (a) Recessive epistasis (gene e is epistatic to gene B) [1 mark]

(b) EeBb produces gametes: EB, Eb, eB, eb. eebb produces gametes: eb only. Offspring genotypes and phenotypes:

EeBb (1/4): Black (E_ B_)
Eebb (1/4): Brown (E_ bb)
eeBb (1/4): Yellow (ee B_)
eebb (1/4): Yellow (ee bb) Phenotypic ratio: 1 black : 1 brown : 2 yellow. [4 marks – 1 for correct gametes, 1 for correct genotypes, 1 for correct phenotype assignment, 1 for correct ratio]

8. (a) Expected ratio for complementary gene action: 9 purple : 7 white [1 mark]

(b) Total F2 = 112 + 88 = 200. Expected purple = 9/16 × 200 = 112.5 Expected white = 7/16 × 200 = 87.5 χ² = (112 − 112.5)²/112.5 + (88 − 87.5)²/87.5 χ² = 0.25/112.5 + 0.25/87.5 χ² = 0.00222 + 0.00286 = 0.00508 (≈ 0.005) [4 marks – 1 for correct total, 1 for correct expected values, 1 for correct χ² formula application, 1 for correct calculation]

(c) Since χ² (0.005) < critical value (3.84), we accept the null hypothesis. The observed results fit the expected 9:7 ratio; the difference is due to chance. [1 mark]

9. (a) Dominant epistasis [1 mark]

(b) WwYy produces gametes: WY, Wy, wY, wy. wwyy produces gametes: wy only. Offspring genotypes and phenotypes:

WwYy (1/4): White (W_ Y_)
Wwyy (1/4): White (W_ yy)
wwYy (1/4): Yellow (ww Y_)
wwyy (1/4): Green (ww yy) Phenotypic ratio: 2 white : 1 yellow : 1 green. [3 marks – 1 for correct gametes, 1 for correct genotypes, 1 for correct phenotypic ratio]

10. (a) Expected F2 phenotypic ratio: 9 agouti : 3 black : 4 albino [1 mark]

(b) The typical 9:3:3:1 ratio is modified because the cc genotype is epistatic to the A gene. When an individual is homozygous recessive cc, no pigment is produced regardless of the A gene alleles, resulting in albino mice. This masks the expression of the A gene, combining the 3 (aaB_) and 1 (aabb) classes with the albino phenotype, giving a 9:3:4 ratio. [2 marks – 1 for identifying epistasis, 1 for explaining masking effect]

Section C: Chromosomal Inheritance and Genetic Analysis (20 marks)

11. (a) I-2 is affected, so must have at least one H allele. Since Huntington's is autosomal dominant and rare, I-2 is most likely heterozygous (Hh). If homozygous (HH), all offspring would be affected, but II-2 is unaffected (hh). Therefore, I-2 must be Hh. [2 marks – 1 for identifying Hh, 1 for reasoning using unaffected offspring]

(b) II-3 is affected (Hh, as his mother I-2 is Hh and father I-1 is hh). II-4 is unaffected (hh). Their son III-1 could inherit the h allele from II-3 (probability 1/2) and h from II-4, resulting in genotype hh (unaffected). [2 marks – 1 for identifying parental genotypes, 1 for explaining inheritance of recessive allele]

(c) III-4 is affected, so genotype is Hh (inherited H from II-3, h from II-4). Unaffected man is hh. Offspring: 1/2 Hh (affected), 1/2 hh (unaffected). Probability of affected child = 1/2 (50%). [2 marks – 1 for correct genotypes, 1 for correct probability]

12. (a) The father is heterozygous: one allele has the normal restriction site (8.0 kb fragment), and the other allele has the additional restriction site, producing 5.0 kb and 3.0 kb fragments. The mother is homozygous for the normal allele (both alleles produce 8.0 kb fragments). [3 marks – 1 for identifying father as heterozygous, 1 for explaining fragment origin, 1 for identifying mother as homozygous normal]

(b) Child 2 shows only the 8.0 kb fragment, indicating homozygosity for the normal allele. Genotype: homozygous normal (unaffected). [2 marks – 1 for correct genotype, 1 for explanation]

(c) The disorder is dominant. The father is affected and heterozygous (one disease allele, one normal allele). Child 2 inherited only normal alleles and is unaffected. If the disorder were recessive, the father would need two disease alleles to be affected, but he shows both normal and disease fragments. [2 marks – 1 for correct answer, 1 for justification using data]

13. (a) Red-eyed female (homozygous): X^R X^R; White-eyed male: X^r Y [2 marks – 1 for each correct genotype]

(b) Female gametes: all X^R. Male gametes: X^r, Y. F1: Females X^R X^r (red-eyed), Males X^R Y (red-eyed). All F1 offspring have red eyes. Ratio: 1 red-eyed female : 1 red-eyed male. [3 marks – 1 for correct gametes, 1 for correct genotypes, 1 for correct phenotypes and ratio]

X^R X^R: red-eyed female (1/4)
X^R X^r: red-eyed female (1/4)
X^R Y: red-eyed male (1/4)
X^r Y: white-eyed male (1/4) Phenotypes: 2 red-eyed females : 1 red-eyed male : 1 white-eyed male. [4 marks – 1 for correct gametes, 1 for correct Punnett square, 1 for correct genotypes, 1 for correct phenotypic ratio]

14. (a) Let X^D = normal allele, X^d = DMD allele. The woman's brother has DMD (X^d Y), so their mother must be a carrier (X^D X^d). The woman is unaffected, so her genotype could be X^D X^D or X^D X^d. Since her mother is a carrier, the woman has a 1/2 chance of being a carrier (X^D X^d). Her husband is unaffected: X^D Y. If the woman is X^D X^D: all offspring unaffected. If the woman is X^D X^d: offspring genotypes: X^D X^D (unaffected female), X^D X^d (carrier female), X^D Y (unaffected male), X^d Y (affected male). [4 marks – 1 for correct symbols, 1 for identifying woman's possible genotypes, 1 for husband's genotype, 1 for correct offspring genotypes]

(b) Probability of affected son = (probability woman is carrier) × (probability of X^d Y son) = 1/2 × 1/4 = 1/8 (12.5%). [1 mark]

(c) An unaffected daughter could inherit X^D from her father and X^d from her mother (if the mother is a carrier). Her genotype would be X^D X^d, making her a carrier. She is unaffected because the normal allele X^D is dominant over the DMD allele X^d. [2 marks – 1 for explaining possible inheritance of X^d, 1 for explaining why she is unaffected despite being a carrier]

15. (a) Green-feathered male (heterozygous): Z^G Z^g; Blue-feathered female: Z^g W [2 marks – 1 for each correct genotype]

(b) Male gametes: Z^G, Z^g. Female gametes: Z^g, W. Offspring:

Z^G Z^g: green male (1/4)
Z^G W: green female (1/4)
Z^g Z^g: blue male (1/4)
Z^g W: blue female (1/4) Phenotypic ratio: 1 green male : 1 green female : 1 blue male : 1 blue female. [3 marks – 1 for correct gametes, 1 for correct genotypes, 1 for correct phenotypic ratio]

Section D: Genetic Variation and Advanced Topics (10 marks)

16. (a) Total population = 360 + 40 = 400. Frequency of white (recessive) phenotype (bb) = q² = 40/400 = 0.1. Frequency of recessive allele (b) = q = √0.1 = 0.316 (or 0.32). [2 marks – 1 for correct q², 1 for correct q]

(b) Frequency of dominant allele (B) = p = 1 − q = 1 − 0.316 = 0.684. Frequency of heterozygotes = 2pq = 2 × 0.684 × 0.316 = 0.432. Number of heterozygous black butterflies = 0.432 × 400 = 172.8 ≈ 173. [3 marks – 1 for correct p, 1 for correct 2pq, 1 for correct number]

17. Crossing over occurs during prophase I of meiosis, when homologous chromosomes pair up and exchange segments of genetic material. This process creates new combinations of alleles on a chromosome (recombinant chromosomes) that were not present in either parent. This increases genetic variation in gametes, as each gamete can contain a unique combination of alleles, contributing to diversity in offspring. [3 marks – 1 for when it occurs, 1 for mechanism of exchange, 1 for effect on genetic variation]

18. Linked genes are located on the same chromosome and tend to be inherited together, as they do not assort independently. Their inheritance pattern shows parental phenotypes more frequently than recombinant phenotypes in test crosses. Unlinked genes are located on different chromosomes (or far apart on the same chromosome) and assort independently according to Mendel's law of independent assortment, producing a 1:1:1:1 ratio of gametes in a dihybrid test cross. [3 marks – 1 for definition of linked genes, 1 for definition of unlinked genes, 1 for comparison of inheritance patterns]

19. The man has blood type AB (genotype I^A I^B). The woman has blood type O (genotype ii). Their children can inherit I^A or I^B from the father and i from the mother, resulting in genotypes I^A i (blood type A) or I^B i (blood type B). No other blood types are possible because the mother can only contribute an i allele, and the father can only contribute I^A or I^B. Blood types AB and O are not possible because they require two specific alleles (I^A I^B or ii) that cannot be produced from this cross. [2 marks – 1 for correct possible blood types, 1 for explanation]

20. Total plants = 500. Entire leaves (recessive phenotype, ll) = 80. Frequency of recessive phenotype (ll) = q² = 80/500 = 0.16. Frequency of recessive allele (l) = q = √0.16 = 0.4. Frequency of dominant allele (L) = p = 1 − 0.4 = 0.6. Frequency of heterozygotes = 2pq = 2 × 0.6 × 0.4 = 0.48. Expected number of heterozygous plants = 0.48 × 500 = 240. [2 marks – 1 for correct allele frequencies, 1 for correct number of heterozygotes]

END OF ANSWER KEY