Mbchb Cat - Genetics Answer Key

## MBMB2211B - Date: 27/3/2025 --- ## SECTION A (30 marks) - MCQs ### Question 1 **Two pure breeding parents are crossed similar to Mendel's Parental generation. A tall plant is crossed with a short plant. What is the expected outcome for the F1 generation?** A. all shortB. all tallC. all medium heightD. half tall, half short **Answer: B. all tall** **Explanation:** When crossing pure breeding tall (TT) with pure breeding short (tt) plants, all F1 offspring will be heterozygous (Tt). Since tall is dominant over short, all F1 plants will express the tall phenotype. --- ### Question 2 **If the cross from #1 is continued, what would be the expected outcome in the F2 generation?** A. all shortB. all tallC. 3 tall, 1 shortD. half tall, half short **Answer: C. 3 tall, 1 short** **Explanation:** F1 × F1 cross (Tt × Tt) produces: 1 TT : 2 Tt : 1 tt genotypic ratio. Since T is dominant, the phenotypic ratio is 3 tall : 1 short. --- ### Question 3 **Which of the following is the BEST definition of inbreeding?** A. mating of individuals who are closely related geneticallyB. mating of genetically unrelated individualsC. mating in which individuals who are alike for particular trait or phenotype mateD. mating in which individuals who are not genetically alike for particular traits mate **Answer: A. mating of individuals who are closely related genetically** **Explanation:** Inbreeding specifically refers to reproduction between genetically related individuals, which increases homozygosity and can lead to expression of deleterious recessive alleles. --- ### Question 4 **If a plant that has round seeds has a parent that has wrinkled seeds, what is its genotype? (Assume that round is dominant.)** A. RRB. RrC. rrD. RrWw **Answer: B. Rr** **Explanation:** Since the plant has round seeds (dominant phenotype) but has a parent with wrinkled seeds (rr), the round-seeded plant must be heterozygous (Rr) to have inherited the recessive allele from the wrinkled parent. --- ### Question 5 **Below is a Punnett square showing a cross between two parents. Use this information to respond to the next three questions. P generation: BB x bb Complete dominance: B = black rat b = white rat** **Referring to the Punnett square above, which of the following accurately represents the phenotypic and genotypic ratios of the F1 generation?** A. Phenotypic ratio 100% white, genotypic ratio 100% BbB. Genotypic ratio 100% black, phenotypic ratio 100% BbC. Phenotypic ratio 100% black, genotypic ratio 100% BbD. Phenotypic ratio 50% black, 50% white, genotypic ratio 100% Bb **Answer: C. Phenotypic ratio 100% black, genotypic ratio 100% Bb** **Explanation:** BB × bb cross produces all Bb offspring. Since B (black) is dominant over b (white), all offspring will be black phenotypically but heterozygous (Bb) genotypically. --- ### Question 6 **Using the information in the Punnett square above, how would we refer to the parents and the offspring?** A. One parent homozygous, one is heterozygous, the offspring are homozygousB. One parent is homozygous dominant, one parent is homozygous recessive, the offspring are heterozygousC. One parent is homozygous dominant, one parent is heterozygous recessive, the offspring are homozygous dominantD. One parent is heterozygous dominant, one is heterozygous recessive, the offspring are heterozygous dominant **Answer: B. One parent is homozygous dominant, one parent is homozygous recessive, the offspring are heterozygous** **Explanation:** BB is homozygous dominant, bb is homozygous recessive, and all Bb offspring are heterozygous. --- ### Question 7 **If we were to cross the offspring in the Punnett square above (known as the F1 generation), what will be the genotypic and phenotypic ratios of the F2 generation?** A. Phenotypic ratio 3:1; genotypic ratio 1:2:1B. Phenotypic ratio 1:2:1, genotypic ratio 3:1C. Phenotypic ratio 3:1, genotypic ratio 3:1D. Phenotypic ratio 1:2:1, genotypic ratio 1:2:1 **Answer: A. Phenotypic ratio 3:1; genotypic ratio 1:2:1** **Explanation:** F1 × F1 (Bb × Bb) produces: 1 BB : 2 Bb : 1 bb (genotypic ratio 1:2:1). Phenotypically: 3 black : 1 white (ratio 3:1). --- ### Question 8 **Recombination frequency of < 0.5 suggests:** A. LinkageB. No linkageC. HomozygotesD. Heterozygotes **Answer: A. Linkage** **Explanation:** Recombination frequency less than 50% indicates that genes are linked on the same chromosome. Independent assortment would give 50% recombination frequency. --- ### Question 9 **Use the information and the Punnett square below to respond to the next three questions: Hemophilia is an X linked recessive disorder** **What are the chances that this couple will have a child with hemophilia?** A. 75%B. 50%C. 25%D. 0% **Answer: C. 25%** **Explanation:** Assuming carrier female (X^H X^h) × normal male (X^H Y): 25% X^H X^H, 25% X^H X^h, 25% X^H Y, 25% X^h Y. Only X^h Y males have hemophilia = 25%. --- ###