Chapter 5
5.1, 5.2, 5.12, 5.13
5.1: The term “inactive” is not quite correct because about 15% of the genes in this chromosome are transcriptionally active to some degree, producing 15−50% of the RNA transcripts as are produced from the active X chromosome.
5.2: The genetic consequence of the obligatory crossover is that alleles in the pseudoautosomal region in the X chromosome can be interchanged with their homologous alleles in the pseudoautosomal region in the Y chromosome, yielding a pattern of inheritance in pedigrees that is indistinguishable from that of ordinary autosomal inheritance.
5.12: The mother has a translocation that joins the long arm of chromosome 21 with the long arm of another acrocentric chromosome (a ‘Robertsonian’ translocation). The affected child has 46 chromosomes. This karyotype differs from the usual karyotype for trisomy 21 because the extra chromosome 21 is not a free chromosome.
5.13: Because the X chromosome in the 45, X daughter contains the color-blindness allele, the 45, X daughter must have received the X chromosome from her father by way of a normal X-bearing sperm. The nondisjunction must, therefore, have occurred in the mother, resulting in an egg cell lacking an X chromosome.
Additional Question not in your textbook:
#1. In Drosophila, waxy wings are inherited as a sex-linked recessive trait, and hairy body is inherited as an autosomal dominant trait. Indicate the F1 and F2 phenotypic ratios expected from a cross between a waxy wing male and a hairy body female.
Approach this problem by first denoting the genotypes at both genes of the parents:
P waxy male X hairy female
wY h+h+ w+w+ HH
Then, proceed to the F1 generation. Remember that males get their X chromosome from their mother and their Y chromosome from the father. Females get an X chromosome from each of the mother and father.
F1 males are: w+Y Hh+
females are: w+w Hh+
Next, indicate the F2 generalized genotypes that will result in each phenotype. Calculate the probability of obtaining the genotype at each gene pair and multiply the individual probabilities. This give the expected ratio of the corresponding phenotype as indicated below:
F2 w+_ H_ probability = ½ X ¾ = 3/8 hairy females
w+_ h+h+ probability = ½ X ¼ = 1/8 wildtype females
w+Y H_ probability = ¼ X ¾ = 3/16 hairy males
w+Y h+h+ probability = ¼ X ¼ = 1/16 wildtype males
wY H_ probability = ¼ X ¾ = 3/16 waxy, hairy males
wY h+h+ probability = ¼ X ¼ = 1/16 waxy males
