HKDSE Biology Practice Paper: Genetics And Inheritance (12 Questions, 100 Marks, 75 Minutes)

HKDSE Biology Practice Paper: Genetics and Inheritance (12 Questions, 100 Marks, 75 Minutes)

2026-11-03

HKDSE Biology: Genetics and Inheritance

HKDSE Biology practice paper on genetics and inheritance: 12 questions, 100 marks, 75 minutes, covering Mendelian crosses, sex linkage, pedigrees, multiple alleles and DNA base composition.

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Paper details　Level: S4-S6 | Subject: Biology | Time: 75 minutes | Marks: 100 | Questions: 12

Section A: Mendelian Genetics and Cross Analysis (Questions 1-4, 32 marks)

Each question carries 8 marks; state the parental gametes, the genotype ratio and the phenotype ratio, then calculate the numbers from the data given.

#	Question	Answer and marking notes
1	In a pea plant, the allele T for tall stem is completely dominant over the allele t for dwarf stem. Two tall plants (Tt × Tt) are crossed and 400 seeds are obtained, all of which grow into plants. (a) State the gametes produced by each parent and their ratio. (b) Deduce the genotype ratio and the phenotype ratio of the offspring. (c) Calculate how many of the 400 offspring are expected to be tall and how many dwarf.	(a) Both parents are Tt and each produces two kinds of gamete, T and t, in a 1:1 ratio. (b) The offspring genotypes are TT:Tt:tt = 1:2:1, that is 25%:50%:25%; because T is completely dominant the phenotype ratio is tall:dwarf = 3:1. (c) 300 tall and 100 dwarf plants are expected (400 × 3/4 and 400 × 1/4). [Marks: gametes 2; genotype ratio 2; phenotype ratio 2; numbers 2.]
2	A tall pea plant, known to be either TT or Tt, is test crossed with a dwarf plant (tt). The 400 offspring contain approximately equal numbers of tall and dwarf plants. (a) Deduce the genotype of the tall parent and explain. (b) State the phenotype ratio of this test cross and the expected number of plants of each phenotype. (c) How would the result differ if the parent were TT?	(a) The tall parent is Tt. Dwarf (tt) offspring appear in the test cross, so the parent must have contributed a t gamete and cannot be TT. (b) Tt × tt gives 50% Tt and 50% tt, a tall:dwarf phenotype ratio of 1:1, so 200 tall and 200 dwarf plants are expected among 400. (c) If the parent were TT, all offspring of TT × tt would be Tt and therefore 100% tall, so no dwarf plants would appear. [Marks: genotype with reason 4; 1:1 and numbers 3; TT case 2.]
3	In peas, T (tall) is completely dominant over t (dwarf) and R (red flowers) over r (white flowers); the two genes are on different pairs of chromosomes. Two double heterozygotes TtRr are crossed and 1,600 seeds are obtained. (a) State the gametes produced by the parents and their ratio. (b) Deduce the phenotype ratio of the offspring. (c) Calculate the expected number of plants of each of the four phenotypes.	(a) A TtRr double heterozygote produces four kinds of gamete, TR, Tr, tR and tr, in a 1:1:1:1 ratio (25% each). (b) Because the two genes assort independently, the offspring phenotype ratio is 9 tall red : 3 tall white : 3 dwarf red : 1 dwarf white. (c) 900 tall red, 300 tall white, 300 dwarf red and 100 dwarf white plants, 1,600 in total. [Marks: gametes 2; 9:3:3:1 with explanation 3; four numbers 3.]
4	A TtRr plant is crossed with a ttrr plant and 800 seeds are obtained. (a) What is this type of cross called and what is its purpose? (b) Deduce the phenotype ratio of the offspring and the number of plants of each phenotype. (c) What is the probability that an offspring is the homozygous tall white-flowered genotype TTrr, and how many such plants are expected here?	(a) It is a test cross, a cross with a double recessive individual, used to determine the genotype of an unknown individual and to reveal its gamete ratio. (b) Under independent assortment the offspring show a 1:1:1:1 phenotype ratio, that is 200 plants of each of tall red, tall white, dwarf red and dwarf white. (c) The homozygous tall white genotype is TTrr and requires two Tr gametes, so the probability is 25% × 25% = 6.25%; among 800 seeds, 50 such plants are expected. [Marks: name and purpose 3; ratio and numbers 3; probability and number 2.]

Section B: Incomplete Dominance, Sex Linkage and Pedigree Probability (Questions 5-8, 36 marks)

Each question carries 9 marks; identify the mode of inheritance (dominant, recessive, sex-linked or probability) and derive every ratio or percentage from the data given.

#	Question	Answer and marking notes
5	In an ornamental plant, flower colour is controlled by one pair of alleles: R (red) and W (white). The heterozygote RW has pink flowers, showing incomplete dominance. Two pink-flowered plants are crossed and 1,200 seeds are obtained. (a) State the gametes of the parents and the genotype ratio of the offspring. (b) State the phenotype ratio and explain how it differs from the case of complete dominance. (c) Calculate the expected number of plants of each flower colour.	(a) RW × RW produces R and W gametes in a 1:1 ratio; the offspring genotypes are RR:RW:WW = 1:2:1. (b) With incomplete dominance the heterozygote shows an intermediate phenotype (pink), so the phenotype ratio is also 1 red : 2 pink : 1 white; with complete dominance the heterozygote would look like the dominant homozygote and the ratio would be 3:1. (c) 300 red, 600 pink and 300 white plants, 1,200 in total. [Marks: genotype ratio 2; phenotype ratio and contrast 3; three numbers 3.]
6	Red-green colour blindness is caused by a recessive allele b on the X chromosome; XB is the normal allele. A carrier mother (XBXb) marries a normal father (XBY) and they expect 400 sons and 400 daughters. (a) State the gametes produced by each parent. (b) Deduce the phenotype distribution among the sons and among the daughters. (c) Calculate how many sons are expected to be colour blind and how many daughters are expected to be carriers.	(a) The mother XBXb produces XB and Xb eggs (1:1); the father XBY produces XB and Y sperm (1:1). (b) Among the sons (XY), 50% are XBY and normal while 50% are XbY and colour blind; among the daughters (XX), 50% are XBXB and 50% are XBXb carriers, and all appear normal. (c) Of 400 sons, 200 are expected to be colour blind; of 400 daughters, 0 are affected and 200 are carriers. [Marks: gametes 2; sons and daughters 4; numbers 3.]
7	A phenotypically normal couple have a first child affected by a rare autosomal recessive disorder (aa). (a) State the genotypes of the parents and explain why both appear normal. (b) Deduce the probability that each child is affected. (c) If they have three more children, what is the probability that all three are phenotypically normal? Give your answer as a percentage to one decimal place.	(a) Both are carriers, Aa. The recessive disorder appears only in aa homozygotes; a carrier has one a and one dominant A, so it remains phenotypically normal. (b) Aa × Aa gives 1 AA : 2 Aa : 1 aa, so the probability that each child is affected (aa) is 1/4, that is 25%. (c) Each child has a 3/4 probability of being normal, so three normal children have probability (3/4)³ = 27/64 ≈ 0.4219, that is 42.2%. [Marks: genotypes and explanation 3; 1/4 2; working 2; percentage 2.]
8	Mendel’s dihybrid experiment yielded 556 seeds with observed numbers of 315, 108, 101 and 32 for the four phenotypes; independent assortment predicts a 9:3:3:1 ratio. (a) Calculate the expected number for each phenotype (to two decimal places) and check the total. (b) Which phenotype shows the largest deviation between the observed and expected numbers? State that deviation. (c) Explain why the observed numbers never equal the expected numbers exactly.	(a) The total is 556 seeds. On a 9:3:3:1 ratio the expected numbers are 556×9/16 = 312.75, 556×3/16 = 104.25, 104.25 and 556×1/16 = 34.75, which total 556.00 and match the observed total. (b) The second phenotype deviates most: observed 108 against expected 104.25, a deviation of +3.75 (the others are +2.25, −3.25 and −2.75). (c) Segregation and independent assortment are probabilistic: gamete formation and fertilisation are random and the sample is finite, so observed numbers fluctuate about the expected values rather than matching them exactly. [Marks: four expectations 4; largest deviation 2; randomness 3.]

Section C: Multiple Alleles, a Lethal Allele and DNA Composition (Questions 9-12, 32 marks)

Each question carries 8 marks; use correct allele symbols, show the working for each calculation, and make the conclusion consistent with the base-pairing rule.

#	Question	Answer and marking notes
9	Human ABO blood groups are controlled by three alleles IA, IB and i; IA and IB are codominant and i is recessive. A group A man (IAi) marries a group B woman (IBi) and they expect 400 children. (a) State the gametes of each parent. (b) Deduce the genotype ratio and the blood group ratio of the children. (c) Calculate the expected number of children of each blood group.	(a) The man IAi produces IA and i gametes and the woman IBi produces IB and i gametes, each 1:1. (b) The children are IAIB:IAi:IBi:ii = 1:1:1:1, that is blood groups AB:A:B:O = 1:1:1:1. (c) Among 400 children, 100 of each of groups AB, A, B and O are expected. [Marks: gametes 2; genotype ratio 2; blood group ratio 1; four numbers 3.]
10	A group AB man (IAIB) marries a group O woman (ii). (a) State the possible blood groups of their children and the ratio. (b) Explain why this couple cannot have a group O child. (c) If they have 200 children, how many of each of group A and group B are expected?	(a) IAIB × ii can give only IAi (group A) or IBi (group B) children in a 1:1 ratio, that is 50% each; groups AB and O cannot occur. (b) Group O requires the genotype ii, which needs an i allele from each parent; the father IAIB has no i allele, so an ii child is impossible. (c) Of 200 children, 100 are group A and 100 are group B. [Marks: blood groups and ratio 3; reason 3; numbers 2.]
11	In mice, coat colour is controlled by the alleles AY (yellow) and a (agouti); AY is dominant over a, but AYAY homozygotes die during embryonic development. Two yellow mice (AYa × AYa) are mated and 300 fertilised eggs are produced. (a) State the genotype ratio of the fertilised eggs. (b) Explain why the ratio of yellow to agouti among the surviving offspring is not 3:1. (c) Calculate the percentage of yellow mice among the surviving offspring (to one decimal place).	(a) The genotype ratio of the fertilised eggs is AYAY:AYa:aa = 1:2:1. (b) AYAY homozygotes die during embryonic development, so only AYa (yellow) and aa (agouti) survive, in a ratio of 2:1 rather than 3:1. (c) Of 300 fertilised eggs, 75 die, leaving 225 survivors of which 150 are yellow, that is 150/225 = 66.7%. [Marks: 1:2:1 and lethal class 3; 2:1 explanation 3; percentage 2.]
12	A double-stranded DNA molecule contains 2,000 nucleotides, of which adenine (A) makes up 22% of all bases. (a) Using the base-pairing rule, state the percentage of each of the four bases. (b) Calculate the number of each base and check the total. (c) If a stretch of this DNA contains 300 base pairs, how many guanine (G) bases are there on its complementary strand?	(a) A = T = 22%, together 44%; the remaining 56% is shared equally by G and C, so G = C = 28%. (b) 440 A, 440 T, 560 G and 560 C, totalling 2,000 bases as stated. (c) 300 base pairs means 600 bases, of which 28% are G, that is 168 across both strands; because the strands are complementary, the number of G on the complementary strand equals the number of C on the original strand, namely 84. [Marks: four percentages 3; four numbers 3; 84 G 2.]

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