Heredity and evolution sample question 2

21; largely sterile, as three copies of each chromosome cannot segregate evenly

Heredity and evolution sample question 3

stabilizing selection, reducing variation around the mean

MCAT · Biological and Biochemical Foundations · Biology

Heredity and evolution

Heredity and evolution is part of Biology in the MCAT Biological and Biochemical Foundations section. The questions below are written in the style of the real test, each with the answer and a worked explanation.

Question types you will see here include hardy-Weinberg calculation, pedigree inference, linkage and map distance and cross ratios and gene interaction.

Common traps: using q-squared as q; calling carrier frequency q rather than 2pq; forgetting p + q = 1.

Sample questions

Question 1
A gardener has a sweet-pea plant with purple flowers, a dominant trait, and wants to know whether it is homozygous. She crosses it with a white-flowered plant, which is homozygous recessive, and among 100 offspring finds 52 purple and 48 white. What is the genotype of the purple parent?
- A.
  Homozygous dominant (PP)
- B.
  Heterozygous (Pp)
- C.
  Homozygous recessive (pp)
- D.
  It cannot be determined without a second cross.
Answer: B. Heterozygous (Pp)
A test cross to a homozygous recessive reveals the gametes of the unknown parent: a roughly 1 : 1 ratio of purple to white offspring means the parent made P and p gametes equally, so it is Pp. A PP parent would give only purple offspring, a pp parent would be white, and the 52 : 48 ratio is close enough to 1 : 1 that no second cross is needed.
Question 2
A diploid grass has 2n = 14. A tetraploid derived from it (4n = 28) is crossed with the diploid. How many chromosomes do the offspring have, and how fertile are they?
- A.
  21; largely sterile, as three copies of each chromosome cannot segregate evenly
- B.
  28; fertile, with the same chromosome number as the tetraploid parent plant
- C.
  35; fertile, combining the tetraploid's somatic set with one diploid gamete
- D.
  42; largely sterile, combining both parents' full somatic sets
Answer: A. 21; largely sterile, as three copies of each chromosome cannot segregate evenly
The tetraploid gives 14-chromosome gametes and the diploid 7-chromosome gametes, so offspring are triploid with 21; unpaired third homologs give unbalanced gametes, which isolates the tetraploid as a new species (A). B adds two tetraploid gametes. C adds a somatic set to a gamete. D adds both somatic sets.
Question 3
Newborns of intermediate birth weight survive better than very light or very heavy newborns. Over many generations, this pattern of survival is expected to cause:
- A.
  directional selection, shifting the mean toward heavier babies
- B.
  disruptive selection, favoring both extremes of birth weight
- C.
  stabilizing selection, increasing variation in birth weight
- D.
  stabilizing selection, reducing variation around the mean
Answer: D. stabilizing selection, reducing variation around the mean
When intermediate phenotypes have the highest fitness, selection acts against both extremes, which is stabilizing selection; it keeps the mean about the same while narrowing the spread. Disruptive selection is the reverse pattern, in which both extremes are favored over the intermediate.
Question 4
Two parents are both heterozygous for an autosomal recessive disorder and both heterozygous for a different, unlinked, fully penetrant autosomal dominant disorder. What is the probability that their child is affected by neither disorder?
- A.
  1/16
- B.
  3/16
- C.
  9/16
- D.
  3/4
Answer: B. 3/16
For the recessive disorder, 3/4 of children are unaffected (AA or Aa); for the dominant disorder with fully penetrant heterozygotes, only the 1/4 who are homozygous recessive (dd) are unaffected. Independent assortment lets the probabilities multiply: 3/4 × 1/4 = 3/16. The value 9/16 wrongly treats both conditions as recessive, and 1/16 wrongly treats both as dominant.

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