end of chapter 7

Réussis tes devoirs et examens dès maintenant avec Quizwiz!

The introduction to this chapter described the search for genes that determine pattern baldness in humans. In 1916, Dorothy Osborn suggested that pattern baldness is a sex-influenced trait (see Chapter 5) that is dominant in males and recessive in females. More research suggested that pattern baldness is an X-linked recessive trait. Would you expect to see independent assortment between genetic markers on the X chromosome and pattern baldness if pattern baldness is sex influenced?

No, it will show linkage with X-linked genetic markers.

In tomatoes, tall (D) is dominant over dwarf (d) and smooth fruit (P) is dominant over pubescent fruit (p), which is covered with fine hairs. A farmer has two tall and smooth tomato plants, which we will call plant A and plant B. The farmer crosses plants A and B with the same dwarf and pubescent plant and obtains the following numbers of progeny: Progeny of: Plant A Plant B Dd Pp 122 2 Dd pp 6 82 dd Pp 4 82 dd pp 124 4 Explain why different proportions of progeny are produced when plant A and plant B are crossed with the same dwarf pubescent plant.

Plant A has the two loci in coupling configuration; Plant B has the two loci in repulsion.

Recombination rates between three loci in corn are shown below. Loci Percent recombination R and W2 17% R and L2 35% W2 and L2 18% What is the order of the genes on the chromosome?

R, W2, L2 The recombination frequency indicates the distance between two genes. The lower the frequency the lower the distance

What is the difference between a genetic map and a physical map?

A genetic map shows recombination frequencies between genes; a physical map shows distances in terms of a physical measurement such as base pairs of DNA sequence

Assume that you wish to determine if two genes are linked or not. You are going to analyze the progeny of a genetic cross to determine this. Which of the following crosses will MOST easily yield offspring

An individual who is heterozygous for both genes is crossed to another individual who is homozygous recessive for both genes.

Why is the frequency of recombinant gametes for two loci always half the frequency of meiotic events that experience a crossover between the two loci when the two are close to each other?

Because crossing over involves only two of the four chromatids of a homologous chromosome pair

What does the term recombination mean?

Generation of new combinations of alleles that are different from the parental combinations.

In cucumbers, heart-shaped leaves (hl) are recessive to normal leaves (Hl) and having numerous fruit spines (ns) is recessive to having few fruit spines (Ns). The genes for leaf shape and for number of spines are located on the same chromosome; findings from mapping experiments indicate that they are 32.6 m.u. apart. A cucumber plant having heart-shaped leaves and numerous spines is crossed with a plant that is homozygous for normal leaves and few spines. The F1 are crossed with plants that have heart-shaped leaves and numerous spines. What percentage of the progeny would be expected to have heart-shaped leaves and numerous spines (hl, ns)?

In cucumbers, heart-shaped leaves (hl) are recessive to normal leaves (Hl) and having numerous fruit spines (ns) is recessive to having few fruit spines (Ns). The genes for leaf shape and for number of spines are located on the same chromosome; findings from mapping experiments indicate that they are 32.6 m.u. apart. A cucumber plant having heart-shaped leaves and numerous spines is crossed with a plant that is homozygous for normal leaves and few spines. The F1 are crossed with plants that have heart-shaped leaves and numerous spines. What percentage of the progeny would be expected to have heart-shaped leaves and numerous spines (hl, ns)?

What are two main causes of recombination?

Independent assortment of chromosomes and crossing over between homologous chromosomes during meiosis

The introduction to this chapter described the search for genes that determine pattern baldness in humans. In 1916, Dorothy Osborn suggested that pattern baldness is a sex-influenced trait (see Chapter 5) that is dominant in males and recessive in females. More research suggested that pattern baldness is an X-linked recessive trait. Would you expect to see independent assortment between genetic markers on the X chromosome and pattern baldness if pattern baldness is X-linked?

It depends on the distance; it will show linkage to some X-linked genetic markers but may assort independently of more distant X-linked genetic markers.

In silk moths (Bombyx mori), red eyes (re) and white-banded wing (wb) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (re+ and wb+); these two genes are on the same chromosome. A moth homozygous for red eyes and white-banded wings is crossed with a moth homozygous for the wild-type traits. The F1 have normal eyes and normal wings. The F1 are crossed with moths that have red eyes and white-banded wings in a testcross. The progeny of this testcross are wild-type eyes, wild-type wings 418 red eyes, wild-type wings 19 wild-type eyes, white-banded wings 16 red eyes, white-banded wings 426 What is the map distance between the genes for red eyes and white-banded wings?

The F1 heterozygote inherited a chromosome with alleles for red eyes and white-banded wings (re wb) from one parent and a chromosome with alleles for wild-type eyes and wild-type wings (re+ wb+) from the other parent. These are therefore the phenotypes of the nonrecombinant progeny, present in the highest numbers. The recombinants are the 19 with red eyes, wild-type wings and 16 with wild-type eyes, white-banded wings. RF = recombinants/total progeny × 100% = (19 + 16)/879 × 100% = 4.0% The distance between the genes is four map units.

Honeybees have haplodiploid sex determination: females are diploid, developing from fertilized eggs, whereas males are haploid, developing from unfertilized eggs. Otto Mackensen studied linkage relations among eight mutations in honeybees (O. Mackensen, 1958, Journal of Heredity 49:99-102). The following table gives the results of two of Mackensen's crosses, including three recessive mutations: cd (cordovan body color), h (hairless), and ch (chartreuse eye color). Only the genotype of the queen is given. Why is the genotype of the male parent not needed for mapping these genes?

The male parent contributes nothing to the genotype or the phenotype of the male progeny. Given the genotype of the female parent, the genotype of the male parent can be deduced from the phenotypes of the female progeny.

In German cockroaches, bulging eyes (bu) are recessive to normal eyes (bu+) and curved wings (cv) are recessive to straight wings (cv+). Both traits are encoded by autosomal genes that are linked. A cockroach has genotype bu+ bu cv+ cv, and the genes are in repulsion. Which of the following sets of genes will be found in the most common gametes produced by this cockroach?

The most common gametes will have (e) bu cv+. Equally common will be gametes that have bu+ cv, not given among the choices. Since these genes are linked, in repulsion, the wild-types alleles are on different chromosomes. Thus, the cockroach has one chromosome with bu+cv and the homologous chromosome with bu cv+.Meiosis always produces nonrecombinant gametes at higher frequencies than recombinants, so gametes bearing bu cv+ will be produced at higher frequencies than (a) or (b), which are the products of recombination. The choices (c) and (d) have two copies of one locus and no copy of the other locus. They violate Mendelian segregation: Each gamete must contain one allele of each locus.

In cucumbers, heart-shaped leaves (hl) are recessive to normal leaves (Hl) and having numerous fruit spines (ns) is recessive to having few fruit spines (Ns). The genes for leaf shape and for number of spines are located on the same chromosome; findings from mapping experiments indicate that they are 32.6 m.u. apart. A cucumber plant having heart-shaped leaves and numerous spines is crossed with a plant that is homozygous for normal leaves and few spines. The F1 are crossed with plants that have heart-shaped leaves and numerous spines. What percentage of the progeny would be expected to have heart-shaped leaves and few spines (hl, Ns)?

The recombinants should total 32.6%, so each recombinant phenotype will be 16.3% of the progeny. Because the F1 inherited a chromosome with heart-shaped leaves and numerous spines (hl ns) from one parent and a chromosome with normal leaves and few spines (Hl Ns) from the other parent, these are the nonrecombinant phenotypes, and together they total 67.4%, or 33.7% each. The two recombinant phenotypes are heart-shaped leaves with few spines (hl Ns) and normal-shaped leaves with numerous spines (Hl ns). Heart-shaped, numerous spines 33.7% Normal-shaped, few spines 33.7% Heart-shaped, few spines 16.3% Normal-shaped, numerous spines 16.3%

In cucumbers, heart-shaped leaves (hl) are recessive to normal leaves (Hl) and having numerous fruit spines (ns) is recessive to having few fruit spines (Ns). The genes for leaf shape and for number of spines are located on the same chromosome; findings from mapping experiments indicate that they are 32.6 m.u. apart. A cucumber plant having heart-shaped leaves and numerous spines is crossed with a plant that is homozygous for normal leaves and few spines. The F1 are crossed with plants that have heart-shaped leaves and numerous spines. What percentage of the progeny would be expected to have normal leaves and few spines (Hl, Ns)?

The recombinants should total 32.6%, so each recombinant phenotype will be 16.3% of the progeny. Because the F1 inherited a chromosome with heart-shaped leaves and numerous spines (hl ns) from one parent and a chromosome with normal leaves and few spines (Hl Ns) from the other parent, these are the nonrecombinant phenotypes, and together they total 67.4%, or 33.7% each. The two recombinant phenotypes are heart-shaped leaves with few spines (hl Ns) and normal-shaped leaves with numerous spines (Hl ns). Heart-shaped, numerous spines 33.7% Normal-shaped, few spines 33.7% Heart-shaped, few spines 16.3% Normal-shaped, numerous spines 16.3%

In cucumbers, heart-shaped leaves (hl) are recessive to normal leaves (Hl) and having numerous fruit spines (ns) is recessive to having few fruit spines (Ns). The genes for leaf shape and for number of spines are located on the same chromosome; findings from mapping experiments indicate that they are 32.6 m.u. apart. A cucumber plant having heart-shaped leaves and numerous spines is crossed with a plant that is homozygous for normal leaves and few spines. The F1 are crossed with plants that have heart-shaped leaves and numerous spines. What percentage of the progeny would be expected to have normal leaves and numerous spines (Hl, ns)?

The recombinants should total 32.6%, so each recombinant phenotype will be 16.3% of the progeny. Because the F1 inherited a chromosome with heart-shaped leaves and numerous spines (hl ns) from one parent and a chromosome with normal leaves and few spines (Hl Ns) from the other parent, these are the nonrecombinant phenotypes, and together they total 67.4%, or 33.7% each. The two recombinant phenotypes are heart-shaped leaves with few spines (hl Ns) and normal-shaped leaves with numerous spines (Hl ns). Heart-shaped, numerous spines 33.7% Normal-shaped, few spines 33.7% Heart-shaped, few spines 16.3% Normal-shaped, numerous spines 16.3%

True or False? Genes in coupling configuration are linked; genes in repulsion are unlinked.

false

Crossing over ____________ the number of recombinant offspring.

increases

(Problem 5) True or False? Genes in coupling configuration produce more wild-type gametes; genes in repulsion produce fewer wild-type gametes.

true

True or False? Two genes in coupling configuration have wild-type alleles for both loci on one chromosome and the mutant alleles for both loci on the homologous chromosome; genes in repulsion have a wild-type allele for one locus and a mutant allele for the other locus on each homologous chromosome.

true


Ensembles d'études connexes

Ch. 14 Cost Allocation, Customer-Profitability Analysis, & Sales-Variance Analysis

View Set

Chapter 1: Introduction to Nursing Informatics: Managing Healthcare Information

View Set

Geology Lab - Rock Identification

View Set

Military Trng A+ final exam study guide

View Set

Federal tax considerations for life insurance and annuities

View Set

CH 2 Strategy Formulation, Execution, and Governance

View Set

Obj-Oriented Analysis & Design - Test 1

View Set