Book Viewer

Short Contents | Full Contents | Other books @ NCBI

Modern Genetic Analysis Chapter 5. Recombination of Genes
Solved Problem

2.

A human pedigree shows people affected with the rare nail-patella syndrome (misshapen nails and kneecaps) and also gives the ABO blood group genotype of each individual. (Note: ABO blood types are determined by three alleles IA, IB, and i. IA determines group A, IB group B, and i group O. The IA/IB heterozygote gives group AB.) Both loci concerned are autosomal. To answer the following questions, study the accompanying pedigree.

a. Is the nail-patella syndrome a dominant or recessive phenotype? Give reasons to support your answer.
b. Is there evidence of linkage between the nail-patella gene and the gene for ABO blood type, as judged from this pedigree? Why or why not?
c. If there is evidence of linkage, draw the alleles on the relevant homologs of the grandparents. If there is no evidence of linkage, draw the alleles on two homologous pairs.
d. According to your model, which descendants represent recombinants?
e. What is the best estimate of RF?
f. If man III-1 marries a normal woman of blood type O, what is the probability that their first child will be blood type B with nail-patella syndrome?
See answer

Solution
a. Nail-patella syndrome is most likely dominant. We are told that it is a rare abnormality, so it is unlikely that the unaffected people marrying into the family carry a presumptive recessive allele for nail-patella syndrome. Let N be the causative allele. Then all people with the syndrome are heterozygotes N/n because all (probably including the grandmother, too) result from a mating to an n/n normal individual. Notice that the syndrome appears in all three successive generationsanother indication of dominant inheritance.
b. There is evidence of linkage. Notice that most of the affected peoplethose that carry the N allelealso carry the IB allele; most likely, these alleles are linked on the same chromosome.
c.

(The grandmother must carry both recessive alleles in order to produce offspring of genotype i/i and n/n.)

d. Notice that the grandparental mating is equivalent to a testcross, so the recombinants in generation II are

whereas all others are nonrecombinants, being either N IB/n i or n i/n i.
e. Notice that the grandparental cross and the first two crosses in generation II are identical and, of course, are all testcrosses. Three of the total 16 progeny are recombinant (II-5, II-8, and III-3). This gives a recombinant frequency of RF = 18.8 percent. (We cannot include the cross of II-6 × II-7, because the progeny cannot be designated as recombinant or not.)
f.

The two parental classes are always equal, and so are the two recombinant classes. Since RF = 18.8, the parental classes must comprise 81.2 percent of the offspring. Hence, the probability that the first child will have nail-patella syndrome and blood type B is 40.6 percent.

See question

© 1999 by W. H. Freeman and Company. All rights reserved.