Heredity and evolution sample question 1

stabilizing selection, reducing variation around the mean

Heredity and evolution sample question 2

Crinkled is dominant, and homozygous crinkled plants die before being counted, leaving 2 heterozygotes to 1 normal.

Heredity and evolution sample question 3

His daughters will inherit a premutation, which can expand to a full mutation when they pass it on, putting their children at risk.

Heredity and evolution sample question 5

Meiosis II in the father; the son has the trait, because both Y chromosomes are copies of the father's Y.

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
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 2
Two crinkled-leaf plants are crossed, and the progeny are 120 crinkled and 60 normal. What is the best explanation?
- A.
  Crinkled is recessive, and both parents were heterozygous, which gives the usual 3:1 ratio.
- B.
  Crinkled shows incomplete dominance, so the 1:2:1 genotype ratio appears directly among the leaf phenotypes.
- C.
  Crinkled is dominant, and homozygous crinkled plants die before being counted, leaving 2 heterozygotes to 1 normal.
- D.
  Two genes assort independently, and the 9:3:3:1 ratio was counted as only two classes.
Answer: C. Crinkled is dominant, and homozygous crinkled plants die before being counted, leaving 2 heterozygotes to 1 normal.
A 2:1 ratio from a cross of two phenotypically identical parents is the signature of a homozygous lethal allele: Aa × Aa gives 1 AA (dies) : 2 Aa (crinkled) : 1 aa (normal). A would give 3:1 with a recessive trait, which cannot show up in both parents' offspring this way; B would give three distinct phenotypes; D would give four classes.
Question 3
In fragile X syndrome, a CGG repeat in the X-linked FMR1 gene is normally 5–44 repeats, a premutation is 55–200, and a full mutation is above 200 and is methylated, silencing the gene. A man has a premutation of 90 repeats and is unaffected. Which prediction about his descendants is best?
- A.
  His sons will inherit the premutation and be affected, because the allele is on the X chromosome.
- B.
  His daughters will inherit a premutation, which can expand to a full mutation when they pass it on, putting their children at risk.
- C.
  His daughters and their children will be at no risk, because a premutation cannot expand to a full mutation.
- D.
  All of his daughters will have a full mutation, because repeats expand most in father-to-daughter transmission of the X chromosome.
Answer: B. His daughters will inherit a premutation, which can expand to a full mutation when they pass it on, putting their children at risk.
A father gives his daughters his X and his sons his Y, so sons escape; the premutation passes to daughters largely unchanged, but premutation alleles can expand to full mutations in maternal meiosis, so grandchildren are at risk. Sons cannot receive the father's X (A), premutations can expand (C), and large expansions occur mainly in maternal, not paternal, transmission (D).
Question 4
A man with blood type AB and a woman with blood type A, whose own mother had type O, have a first child with type A. What is the probability that their second child has type A or AB?
- A.
  1/4
- B.
  1/2
- C.
  2/3
- D.
  3/4
Answer: D. 3/4
D is correct: the woman is IAi (her mother gave i), so children are IAIA, IAi, IAIB or IBi, each 1/4, and types A or AB make up 3/4; the first child does not change this. A is the chance of type B only; B is the chance of type A only; C wrongly conditions on the first child being A.
Question 5
A man with a Y-linked trait has a 47,XYY son whose mother is unaffected. At which stage did the nondisjunction most likely occur, and what is the son's phenotype for the trait?
- A.
  Meiosis I in the father; the son lacks the trait, because homologous X and Y chromosomes were separated into different cells.
- B.
  Meiosis II in the father; the son has the trait, because both Y chromosomes are copies of the father's Y.
- C.
  Meiosis I in the mother; the son has the trait, because the extra chromosome is a Y.
- D.
  Meiosis II in the mother; the son lacks the trait, because only the father passes a Y.
Answer: B. Meiosis II in the father; the son has the trait, because both Y chromosomes are copies of the father's Y.
A sperm with two Y chromosomes arises when sister chromatids fail to separate in paternal meiosis II; both Ys are copies of the father's Y, so the son carries and expresses the trait. Failure in meiosis I would give an XY sperm and a 47,XXY son (A), and the mother carries no Y (C, D).

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