Chapter 3: Problem 11
What is the basis for homology among chromosomes?
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 3: Problem 11
What is the basis for homology among chromosomes?
All the tools & learning materials you need for study success - in one app.
Get started for free
Thalassemia is an inherited anemic disorder in humans. Affected individuals exhibit either a minor anemia or a major anemia. Assuming that only a single gene pair and two alleles are involved in the inheritance of these conditions, is thalassemia a dominant or recessive disorder?
Two organisms, \(A A B B C C D D E E\) and aabbccddee, are mated to produce an \(\mathrm{F}_{1}\) that is self-fertilized. If the capital letters represent dominant, independently assorting alleles: (a) How many different genotypes will occur in the \(\mathrm{F}_{2}\) ? (b) What proportion of the \(\mathrm{F}_{2}\) genotypes will be recessive for all five loci? (c) Would you change your answers to (a) and/or (b) if the initial cross occurred between \(A A b b C C\)ddee\(\times\)aaBBccDDEE parents? (d) Would you change your answers to (a) and/or (b) if the initial cross occurred between \(A A B B C C D D E E \times\) aabbccddEE parents?
Albinism in humans is inherited as a simple recessive trait. For the following families, determine the genotypes of the parents and offspring. (When two alternative genotypes are possible, list both.) (a) Two normal parents have five children, four normal and one albino. (b) A normal male and an albino female have six children, all normal. (c) A normal male and an albino female have six children, three normal and three albino. (d) Construct a pedigree of the families in (b) and (c). Assume that one of the normal children in (b) and one of the albino children in (c) become the parents of eight children. Add these children to the pedigree, predicting their phenotypes (normal or albino).
An alternative to using the expanded binomial equation and Pascal's triangle in determining probabilities of phenotypes in a subsequent generation when the parents' genotypes are known is to use the following equation: \(\frac{n !}{s ! t !} a^{s} b^{t}\) where \(n\) is the total number of offspring, \(s\) is the number of offspring in one phenotypic category, \(t\) is the number of offspring in the other phenotypic category, \(a\) is the probability of occurrence of the first phenotype, and \(b\) is the probability of the second phenotype. Using this equation, determine the probability of a family of 5 offspring having exactly 2 children afflicted with sickle-cell anemia (an autosomal recessive disease \()\) when both parents are heterozygous for the sickle-cell allele.
A certain type of congenital deafness in humans is caused by a rare autosomal (not X-linked) dominant gene. (a) In a mating involving a deaf man and a deaf woman (both heterozygous), would you expect all the children to be deaf? Explain your answer. (b) In a mating involving a deaf man and a deaf woman (both heterozygous), could all the children have normal hearing? Explain your answer. (c) Another form of deafness is caused by a rare autosomal recessive gene. In a mating involving a deaf man and a deaf woman, could some of the children have normal hearing? Explain your answer.
What do you think about this solution?
We value your feedback to improve our textbook solutions.