Mbchb Cat - Genetics Answer Key – 48 MCQs | Kenya MBChB

48 Year 2: Molecular Genetics and Cytogenetics exam questions on Mbchb Cat - Genetics Answer Key for medical students. Includes MCQs, answers, explanations and

This MCQ set contains 48 questions on Mbchb Cat - Genetics Answer Key in the Year 2: Molecular Genetics and Cytogenetics unit. Each question includes the correct answer and a detailed explanation for active recall and exam preparation.

Q1: 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?

  1. A. all short
  2. B. all tall
  3. C. all medium height
  4. D. half tall, half short

Correct answer: B – all tall

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. ---

Q2: If the cross from 1 is continued, what would be the expected outcome in the F2 generation?

  1. A. all short
  2. B. all tall
  3. C. 3 tall, 1 short
  4. D. half tall, half short

Correct answer: C – 3 tall, 1 short

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. ---

Q3: Which of the following is the BEST definition of inbreeding?

  1. A. mating of individuals who are closely related genetically
  2. B. mating of genetically unrelated individuals
  3. C. mating in which individuals who are alike for particular trait or phenotype mate
  4. D. mating in which individuals who are not genetically alike for particular traits mate

Correct answer: A – mating of individuals who are closely related genetically

Inbreeding specifically refers to reproduction between genetically related individuals, which increases homozygosity and can lead to expression of deleterious recessive alleles. ---

Q4: 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.)

  1. A. RR
  2. B. Rr
  3. C. rr
  4. D. RrWw
  5. E. Rr
  6. F. 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 p

Correct answer: B – Rr

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. ---

Q5: 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 gen

  1. A. Phenotypic ratio 100% white, genotypic ratio 100% Bb
  2. B. Genotypic ratio 100% black, phenotypic ratio 100% Bb
  3. C. Phenotypic ratio 100% black, genotypic ratio 100% Bb
  4. D. Phenotypic ratio 50% black, 50% white, genotypic ratio 100% Bb
  5. E. , all offspring will be black phenotypically but heterozygous (B

Correct answer: C – Phenotypic ratio 100% black, genotypic ratio 100% Bb

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. ---

Q6: Using the information in the Punnett square above, how would we refer to the parents and the offspring?

  1. A. One parent homozygous, one is heterozygous, the offspring are homozygous
  2. B. One parent is homozygous dominant, one parent is homozygous recessive, the offspring are heterozygous
  3. C. One parent is homozygous dominant, one parent is heterozygous recessive, the offspring are homozygous dominant
  4. D. One parent is heterozygous dominant, one is heterozygous recessive, the offspring are heterozygous dominant

Correct answer: B – One parent is homozygous dominant, one parent is homozygous recessive, the offspring are heterozygous

BB is homozygous dominant, bb is homozygous recessive, and all Bb offspring are heterozygous. ---

Q7: 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?

  1. A. Phenotypic ratio 3:1; genotypic ratio 1:2:1
  2. B. Phenotypic ratio 1:2:1, genotypic ratio 3:1
  3. C. Phenotypic ratio 3:1, genotypic ratio 3:1
  4. D. Phenotypic ratio 1:2:1, genotypic ratio 1:2:1
  5. E. produces: 1 BB : 2 Bb : 1 bb (genotypic ratio 1:2:1). Phenotypically: 3 black : 1 white (ratio 3:1).

Correct answer: A – Phenotypic ratio 3:1; genotypic ratio 1:2:1

F1 × F1 (Bb × Bb) produces: 1 BB : 2 Bb : 1 bb (genotypic ratio 1:2:1). Phenotypically: 3 black : 1 white (ratio 3:1). ---

Q8: Recombination frequency of < 0.5 suggests

  1. A. Linkage
  2. B. No linkage
  3. C. Homozygotes
  4. D. Heterozygotes
  5. E. Independent assortment would give 50% recombination frequency.

Correct answer: A – Linkage

Recombination frequency less than 50% indicates that genes are linked on the same chromosome. Independent assortment would give 50% recombination frequency. ---

Q9: 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?

  1. A. 75%
  2. B. 50%
  3. C. 25%
  4. D. 0%

Correct answer: C – 25%

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%. ---

Q10: The mother undergoes amniocentesis during her pregnancy, and is told that the child she is carrying is a boy. What are the chances that her son will have hemophilia?

  1. A. 75%
  2. B. 50%
  3. C. 25%
  4. D. 0%
  5. E. are equally likely, so 50% chanc

Correct answer: B – 50%

Given the child is male, only male offspring are considered: X^H Y (normal) and X^h Y (hemophiliac) are equally likely, so 50% chance. ---

Q11: The mother has a second pregnancy four years later. She is told that her baby will be a girl. What are the chances that her daughter will have hemophilia?

  1. A. 75%
  2. B. 50%
  3. C. 25%
  4. D. 0%

Correct answer: D – 0%

Female offspring are X^H X^H or X^H X^h. Since hemophilia is X-linked recessive and father contributes X^H, no daughters can have hemophilia (they need X^h X^h). ---

Q12: The principle that states that alleles separate during gamete formation is the

  1. A. principle of multiple alleles
  2. B. principle of dominance and recessiveness
  3. C. principle of independent assortment
  4. D. principle of segregation

Correct answer: D – principle of segregation

Mendel's first law (principle of segregation) states that paired alleles separate during gamete formation, with each gamete receiving only one allele. ---

Q13: If you crossed two heterozygous plants, how many of the offspring will also be heterozygous?

  1. A. All
  2. B. Half
  3. C. 3/4
  4. D. 1/4
  5. E. Therefore, 2/4 = 1/2 = 50% are heterozygous.

Correct answer: B – Half

Aa × Aa cross produces: 1 AA : 2 Aa : 1 aa. Therefore, 2/4 = 1/2 = 50% are heterozygous. ---

Q14: Question 14 A plant that has purple flowers is crossed with one that has white flowers. The offspring were half white and half purple. What were the genotypes of the parents?

  1. A. Pp × pp
  2. B. Pp × Pp
  3. C. PP × pp
  4. D. PP × WW
  5. E. Pp × pp

Correct answer: A – Pp × pp

1:1 ratio indicates a testcross between heterozygous (Pp) and homozygous recessive (pp) parents. ---

Q15: Cystic fibrosis is

  1. A. Sex-linked recessive disorder
  2. B. Autosomal dominant disorder
  3. C. Autosomal recessive disorder
  4. D. Sex-linked dominant disorder
  5. E. and requires two copies of the defective allel

Correct answer: C – Autosomal recessive disorder

Cystic fibrosis is caused by mutations in the CFTR gene on chromosome 7 (autosome) and requires two copies of the defective allele. ---

Q16: A dihybrid cross (AaBb × AaBb) will result in what offspring ratio?

  1. A. 4:4:2:2
  2. B. 1:3:1
  3. C. 3:1
  4. D. 9:3:3:1

Correct answer: D – 9:3:3:1

Classic dihybrid cross phenotypic ratio: 9 dominant for both traits : 3 dominant for first, recessive for second : 3 recessive for first, dominant for second : 1 recessive for both. ---

Q17: Phenotype refers to an organism's

  1. A. genetic code
  2. B. physical appearance
  3. C. parents
  4. D. ratio

Correct answer: B – physical appearance

Phenotype is the observable physical or biochemical characteristics of an organism, determined by both genotype and environment. ---

Q18: In each case where Mendel crossed true breeding plants as parents, the offspring displayed only one of the two traits seen in the parents. This observation supports which principle of genetics?

  1. A. Segregation
  2. B. independent assortment
  3. C. dominance and recessiveness
  4. D. allele frequency
  5. E. .

Correct answer: C – dominance and recessiveness

The observation that F1 offspring show only one parental trait demonstrates that one allele (dominant) masks the expression of another (recessive). ---

Q19: A human being has autosomes and sex chromosomes

  1. A. 22 pairs, 1 pair
  2. B. 23, 1
  3. C. 2, 2
  4. D. 23, 23

Correct answer: A – 22 pairs, 1 pair

Humans have 23 pairs of chromosomes total: 22 pairs of autosomes and 1 pair of sex chromosomes (XX or XY). ---

Q20: Somatic cells of a human have chromosomes and are called

  1. A. 10, haploid
  2. B. 92, diploid
  3. C. 23, haploid
  4. D. 46, diploid

Correct answer: D – 46, diploid

Human somatic cells contain 46 chromosomes (23 pairs) and are diploid (2n), meaning they have two copies of each chromosome. ---

Q21: A person who receives an extra chromosome could have

  1. A. Heightened intelligence
  2. B. Red eyes
  3. C. Down Syndrome
  4. D. Polygenic traits

Correct answer: C – Down Syndrome

Down syndrome is caused by trisomy 21 (an extra copy of chromosome 21), making it a classic example of chromosomal abnormality due to extra chromosome. ---

Q22: A selection type which works by constantly removing individuals from both ends of a phenotype distribution thus maintaining the same mean over time is

  1. A. Stabilizing
  2. B. Disruptive
  3. C. Directional
  4. D. None of the above

Correct answer: A – Stabilizing

Stabilizing selection favors intermediate phenotypes and removes extreme variants, maintaining the population mean and reducing variation. ---

Q23: Which selection type fits sickle cell anemia?

  1. A. Stabilizing
  2. B. Disruptive
  3. C. Directional
  4. D. None of the above

Correct answer: A – Stabilizing

In malaria-endemic areas, heterozygotes (HbAS) have selective advantage over both homozygotes, representing balancing/stabilizing selection. ---

Q24: Which selection type fits CCR5 mutant gene?

  1. A. Stabilizing
  2. B. Disruptive
  3. C. Directional
  4. D. None of the above

Correct answer: C – Directional

CCR5-Δ32 mutation provides HIV resistance, leading to directional selection favoring the mutant allele in populations exposed to HIV. ---

Q25: What is an allele?

  1. A. Characteristics of an organism
  2. B. Alternate forms of genes
  3. C. Homologous chromosomes
  4. D. Pair of centrioles

Correct answer: B – Alternate forms of genes

Alleles are different versions of the same gene that occupy the same locus on homologous chromosomes. ---

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