/*! 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} Problem 19 In maize, the genes \(P l\) for ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 19

In maize, the genes \(P l\) for purple leaves (dominant over \(p /\) for green leaves), \(s m\) for salmon silk (recessive to \(S m\) for yellow sills, and \(p y\) for pigmy plant (recessive to \(P\) ) for normal-size plant) are on chromosome \(6,\) with map positions as shown: \\[ \frac{p l}{45} \quad \frac{s m}{55} \quad 65 \\] Hybrids from the cross \(P l s m p y / P l s m p y \times p l S m P y / p l \operatorname{Sm} P y\) were testcrossed with \(p l\) sm \(p y / p l\) sm py plants. Predict the phenotypes of the offspring and their frequencies assuming (a) no interference and (b) complete interference.

Short Answer

Expert verified
(a) Parental: 80%, single crossover: 18%, double crossover: 2%. (b) Parental: 80%, single crossover: 20%, no double crossover.

Step by step solution

01

Understand Gene Linkage and Map Units

The problem involves three genes located on chromosome 6: \(Pl\), \(sm\), and \(py\). The genes are located at 45, 55, and 65 map units, respectively. The distances on a genetic map show the probability of crossover events occurring between these genes. To determine the phenotypes of the offspring, we need to consider the probability of recombination based on these distances.
02

Calculate Recombination Frequencies

The distance between \(Pl\) and \(sm\) is 10 map units (55-45), which gives a recombination frequency of 10%. The distance between \(sm\) and \(py\) is also 10 map units (65-55), with a recombination frequency of 10%. The distance between \(Pl\) and \(py\) is 20 map units (65-45), resulting in a recombination frequency of 20%.
03

Determine Phenotype Frequencies Without Interference

With no interference (a), crossover events between each pair of genes are independent. We calculate the probability for each possible recombination event. Parental types (no recombination) are 1 - 0.2 = 0.8, i.e., 80%. For single crossovers between \(Pl\) and \(sm\) or \(sm\) and \(py\), each is 10%, and double crossovers (between both intervals simultaneously) are \(0.1 \times 0.1 = 0.01\), or 1%. Thus, expected phenotype ratios are: parental types = 80%, single crossovers = 18%, and double crossovers = 2% (split between the two double recombinant phenotypes).
04

Determine Phenotype Frequencies With Complete Interference

Complete interference (b) implies that if one crossover occurs, no other crossovers can occur nearby. Thus, only one crossover event can happen between either \(Pl-sm\) or \(sm-py\). The probability of a recombination event in either region singly is 10% each, meaning we have 20% recombination in total spread over the two intervals, without double crossovers. Therefore, parental phenotypes will still be expected in 80% of the offspring, and single crossover recombinants can be expected in 20% (split evenly as 10% for \(Pl-sm\) and 10% for \(sm-py\)). Double crossover types will not occur due to complete interference.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Recombination Frequency
Recombination frequency is a measure that tells us how often genes on the same chromosome swap alleles due to crossover events during meiosis. If genes are located close to each other, they're less likely to be separated by recombination, and vice versa. This frequency helps us predict genetic outcomes and observe inheritance patterns.

In the context of the maize genes discussed in the problem, recombination frequencies are crucial for determining offspring phenotypes. For example, genes 10 map units apart have a 10% chance of crossover鈥攎eaning 10% of offspring will have new allele combinations from one crossover between these genes.
  • To find the recombination frequency, it's calculated as the percent of offspring showing recombinant traits, which translates to map units directly.
  • If two genes are 20 map units apart, they have a recombination frequency of 20%, indicating a higher likelihood of crossover compared to those only 10 map units apart.
Understanding recombination frequency allows geneticists to create maps of chromosomes reflecting these crossover probabilities. This not only aids in predicting genetic outcomes but also helps in gene mapping efforts.
Chromosomal Interference
Chromosomal interference refers to the phenomenon where one crossover event on a chromosome can inhibit another from occurring nearby. This affects the recombination frequency and thus the prediction of genetic traits by altering the expected frequency of double crossover events.

Two main types are discussed often:
  • No interference: Crossovers occur independently, meaning the occurrence of one does not affect another, allowing us to expect certain overlaps in crossover regions.
  • Complete interference: When the occurrence of one crossover prevents another from happening close by, causing a lack of double crossovers. This results in modified patterns of gene inheritance.
In maize, as the problem outlines, complete interference would alter expected phenotypes significantly by limiting crossover events to single instances in specified intervals, which avoids any crossing over at two intervals at the same time. This is important for accurately predicting the spread of characteristics in offspring.
Gene Mapping
Gene mapping is a technique used to determine the specific locations of genes on a chromosome. It helps us understand the distances between genes and their relative positions, allowing for the prediction of recombination frequencies.

The process involves studying recombination frequencies and mapping distances in map units. These units reflect the probability of recombination between genes, helping researchers map out gene positions on a chromosome accurately.
  • Maps established this way are called genetic maps and are key tools in genetic research.
  • They're used to pinpoint locations of genes, studying genetic disorders, and plant and animal breeding for desired traits.
In our maize case, gene mapping shows locations at 45, 55, and 65 map units for the genes, providing a visual layout that aids in calculating recombination likely to occur between these specific gene locations. This foundational knowledge is essential for understanding genetic processes and creating strategies for breeding and research.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the following testcross, genes \(a\) and \(b\) are \(20 \mathrm{cM}\) apart, and genes \(b\) and \(c\) are \(10 \mathrm{cM}\) apart: \(a+c /+b+\times a b c / a b c\) If the coefficient of coincidence is 0.5 over this interval on the linkage map, how many triply homozygous recessive individuals are expected among 1000 progeny?

If two loci are \(10 \mathrm{cM}\) apart, what proportion of the cells in prophase of the first meiotic division will contain a single crossover in the region between them?

A Drosopbila second chromosome that carried a recessive lethal mutation, \(l(2) g 14,\) was maintained in a stock with a balancer chromosome marked with a dominant mutation for curly wings. This latter mutation, denoted \(C y,\) is also associated with a recessive lethal effect-but this effect is different from that of \(l(2) g 14 .\) Thus, \(l(2) g I 4 / C y\) flies survive, and they have curly wings. Flies without the \(C y\) mutation have straight wings. A researcher crossed \(l(2) g 14 / C y\) females to males that carried second chromosomes with different deletions (all homozygous lethal) balanced over the \(C y\) chromosome (genotype \(D f\) / \(C\) ) . Each cross was scored for the presence or absence of progeny with straight wings.

Genes \(a\) and \(b\) are \(20 \mathrm{cM}\) apart. An \(a^{+} b^{+} / a^{+} b^{+}\) individual was mated with an \(a\) b/a \(b\) individual. (a) Diagram the cross and show the gametes produced by each parent and the genotype of the \(\mathrm{F}_{1}\) (b) What gametes can the \(F_{1}\) produce, and in what proportions? (c) If the \(F_{1}\) was crossed to \(a\) b/a \(b\) individuals, what offspring would be expected, and in what proportions? (d) Is this an example of the coupling or repulsion linkage phase? (c) If the \(F_{1}\) were intercrossed, what offspring would be expected and in what proportions?

Consider a female Drosopbila with the following \(\bar{X}\) chromosome genotype: \\[ \frac{w}{w^{+} \quad d o r^{*}} \\] The recessive alleles \(w\) and dor cause mutant eye colors (white and deep orange, respectively). However, \(w\) is epistatic over dor; that is, the genotypes \(w\) dor/ \(Y\) and \(w\) dor/uv dor have white eyes. If there is 40 percent recombination between \(w\) and \(d o r\), what proportion of the sons from this heterozygous female will show a mutant phenotype? What proportion will have either red or deep orange eyes?
See all solutions

Recommended explanations on Biology Textbooks

Substance Exchange

Read Explanation

Heredity

Read Explanation

Genetic Information

Read Explanation

Microbiology

Read Explanation

Cells

Read Explanation

Biology Experiments

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.