NCEA Level 2 Biology 91157 Genetics free practice questions with answers

Question 1 (21 marks) [Inheritance of one and two genes]

(a) In the snapdragon, flower colour is controlled by one gene showing incomplete dominance: red (RR) and white (rr) are the homozygous phenotypes and the heterozygote (Rr) is pink. Use the allele letters R and r. A snapdragon with genotype Rr is crossed with a snapdragon with genotype Rr. (i) Complete a Punnett square for this cross. (ii) State the genotype ratio and the phenotype ratio of the offspring. (iii) Calculate the probability that two offspring produced by this cross will both have the pink phenotype.

(a) Answer: (ii) genotypes 1 : 2 : 1; phenotypes 2 : 1 : 1; (iii)

• Evidence level: Merit (list the gametes from each parent, complete the Punnett square, then read off the ratios).

• (i) Gametes: R and r from the first parent and R and r from the second parent. The square shows the offspring genotypes RR, Rr, Rr, rr.

• (ii) Genotype ratio 1 : 2 : 1 (RR : Rr : rr); phenotype ratio 2 : 1 : 1 (pink : red : white).

• (iii) P(one pink offspring) =

(b) The ABO blood group in humans is controlled by one gene with three alleles: Iᴬ (A) and Iᴮ (B) are codominant, and both are dominant over i (O). A man with blood group AB (genotype IᴬIᴮ) and a woman with blood group B (genotype Iᴮi) have a child. (i) Complete a Punnett square for this cross. (ii) State all the possible blood groups of the child and the probability of each. (iii) Explain what is meant by ‘codominant’ and why a person with genotype IᴬIᴮ has blood group AB.

(b) Answer: (ii) A

• Evidence level: Merit (gametes → Punnett square → genotypes → blood groups (Iᴬ and Iᴮ both dominant to i)).

• (i) Gametes: Iᴬ or Iᴮ from the man and Iᴮ or i from the woman.

• (ii) group A:

• Achievement: (iii) States that both alleles are expressed.

• Merit: Links to the phenotype: in the heterozygote IᴬIᴮ neither allele masks the other.

• Excellence: Full explanation: codominant alleles are both fully expressed in the heterozygote; Iᴬ codes for the A antigen and Iᴮ for the B antigen, so a person with IᴬIᴮ makes both antigens on their red blood cells and has blood group AB, rather than a blend or only one antigen. 4 1 2 , so P(both) = 1 2 × 1 2 = 1 4 . 1 4 ; AB 1 4 ; B 1 2 ; (iii) both alleles expressed 1 4 ; group AB: 1 4 ; group B: 1 2 .

(c) In the fruit fly, wing length (long dominant over vestigial) and body colour (grey dominant over ebony) are controlled by two genes on different chromosomes. Use the letters V/v for wing length and E/e for body colour. A fruit fly with genotype VvEe is crossed with a fruit fly with genotype vvee. (i) Write the genotypes of the gametes produced by each parent, and complete a Punnett square. (ii) State the phenotype ratio of the offspring. (iii) 1600 offspring are produced. Calculate the expected number with the phenotype ‘long, grey’. (iv) Explain why the genes must be on different chromosomes for this ratio to occur.

(c) Answer: (ii) 1 : 1 : 1 : 1; (iii) 400

• Evidence level: Excellence (gametes (one allele of each gene), Punnett square, phenotype ratio, expected number).

• (i) First parent VvEe gametes: VE, Ve, vE, ve. Second parent vvee gametes: ve.