Supplemental Lecture (97/06/10 update) by Stephen T. Abedon (

  1. Chapter title: Pedigree Analysis Problems
    1. A list of vocabulary words is found toward the end of this document
  2. Vocabulary
    1. No entry.
  3. Practice questions - The below questions are from pp. 60, 68-70 of Sinnott et al., 1958:
    1. Which best describes the genetics of the afflicting allele in the following pedigree (it is a pedigree of taste blindness)? [PEEK]
      1. autosomal dominant
      2. autosomal recessive
      3. X-linked dominant
      4. X-linked recessive
      5. Y-linked dominant
      6. Y-linked recessive

    2. Albinism is inherited as an autosomal recessive. In the figure below, assuming that persons from the general population are not heterozygous for albinism (Aa), what are the genotypes of all persons whose genotypes are known? (i.e., indicate the genotypes on the figure for all known AA, Aa, and aa individuals) [PEEK]
    3. A blue-eyed man marries a brown-eyed woman. They have one child, who is blue eyed. What are the genotypes of all the individuals mentioned? [PEEK]
    4. Given the below pedigree, would you expect to find more of in Cleopatra-Berenike III compared with the general population? (figure from p. 283 of R. Lewis, 1998, Life Third Edition. McGraw Hill, Boston, Mass.). [PEEK]
      1. Loci which are heterozygous
      2. Loci which are homozygous for rare alleles
      3. Loci which display epistasis
      4. Loci which display codominance
      5. Alleles
      6. Loci

    5. Suppose that an allele, b, of a sex-linked gene is recessive and lethal. A man marries a woman who is heterozygous for this gene. If this couple had a large number of normal children, what would be the predicted sex ratio of these children (ratio of male children to female children)? (adapted from J.L. Gould and W.T. Keeton (1996). Biological Science. Sixth Edition. W.W. Norton & Company. New York. P. 443) [PEEK]
    6. In Drosophila melanogaster there is a dominant allele for gray body color and a dominant allele of another gene for normal wings. The recessive alleles of these two genes result in black body color and vestigial wings, respectively. Flies homozygous for gray body and normal wings are crossed with flies that have black bodies and vestigial wings. The F1 progeny are then crossed, with the following results:
    7. Gray body, normal wings: 236
    8. Black body, vestigial wings: 253
    9. Gray body, vestigial wings: 50
    10. Black body, normal wings: 61
    11. Would you say that these two genes are linked? If so, how many map units apart are they on the linkage map? (adapted from J.L. Gould and W.T. Keeton (1996). Biological Science. Sixth Edition. W.W. Norton & Company. New York. P. 443) [PEEK]
    12. An afflicted woman marries an afflicted man. The allele causing the affliction is recessive to the wild type allele. What fraction of their children will be silent carriers? What fraction will be afflicted? [PEEK]
    13. For the following pedigree, how does the afflicting allele impact on phenotype? (figure is from M.R Cummings (1988). Human Heredity: Principles and Issues West Publishing Company. New York. p. 113) [PEEK]
      1. it displays complete dominance
      2. it is recessive to the non-afflicting allele otherwise found at the same loci
      3. it displays epistasis
      4. it displays pleiotropic effects
      5. it produces a tRNA product
      6. it is a Mendelian ratio

    14. If a husband and wife have four daughters. Under most circumstances, what is the probability that their fifth child will also be a girl? (adapted from G.T. Licata & W.H. Garnsey (1986). A General Review of Biology 2nd Edition. N & N Publishing, Middletown, NY, p. 119 [PEEK]
    15. A male cat has short hair, a stubby tail, and extra toes. A female cat has long hair, a long tail, and extra toes. The genes and alleles are as follows:
    16. L = short hair; l = long hair
    17. M = stubby tail; m = long tail
    18. P = extra toes; p = normal number of toes
    19. The two cats have kittens. One kitten has long hair, a long tail, and no extra toes. Another has short hair, a stubby tail, and extra toes. The third kitten has short hair, a long tail, and no extra toes. What is the genotype of the father. (adapted from R. Lewis (1998). Life Third Edition. McGraw-Hill, Boston, Mass., p. 283)[PEEK]
    20. The recombination frequency between linked genes A and B is 40%, between B and C is 20%, between C and D is 10%, between C and A is 20%, and between D and B is 10%. What is the sequence of the genes on the chromosome? (adapted from J.L. Gould and W.T. Keeton (1996). Biological Science. Sixth Edition. W.W. Norton & Company. New York. P. 443)[PEEK]
  4. Practice question answers
    1. ii, autosomal recessive.
    2. in the figure below, all individuals from the general population (i.e., married in spouses) are AA. All circled individuals are Aa. All afflicted individuals are aa.
    3. if b is the recessive blue-eyed allele and B the dominant brown-eyed allele, then the man is bb, the woman Bb, and the child bb.
    4. ii, Loci which are homozygous for rare alleles
    5. 1:2 (i.e., half the male children would die)
    6. (50 +61) / (236 + 253 + 50 +61) = 18.5, so yes and 18.5 map units apart
    7. All will be afflicted and none will be silent carriers.
    8. i, it displays complete dominance.
    9. 0.5
    10. Ll Mm Pp
    11. A-C-D-B
  5. References
    1. Sinnott, E.W., Dunn, L.C., Dobzhansky, T. (1958). Principles of Genetics. Fifth Edition. McGraw-Hill Book Co., Inc. New York. p. 60, 68-70.