/*! 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 42 A rice breeder obtained a triple... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A rice breeder obtained a triple heterozygote carrying the three recessive alleles for albino flowers \((a l),\) brown awns \((b),\) and fuzzy leaves \((f u),\) all paired with their normal wild-type alleles. This triple heterozygote was testcrossed. The progeny phenotypes were \(170 \quad\) wild type \(710 \quad\) albino \(150 \quad\) albino, brown, fuzzy \(698 \quad\) brown, fuzzy 5 brown 42 fuzzy \(3 \quad\) albino, fuzzy \(38 \quad\) albino, brown a. Are any of the genes linked? If so, draw a map labeled with map distances. (Don't bother with a correction for multiple crossovers.) b. The triple heterozygote was originally made by crossing two pure lines. What were their genotypes?

Short Answer

Expert verified
Yes, genes are linked. The map distances are: albino-brown 4.4%, brown-fuzzy 4.5%, albino-fuzzy 0.5%. The genotypes were (albino, brown, fuzzy) x (wild-type).

Step by step solution

01

Understanding Linkage

If genes are linked, they do not assort independently and will produce non-Mendelian phenotypic ratios among progeny. We determine linkage by looking for non-random associations among alleles.
02

Identify Parent Phenotypes

The wild-type phenotype and the phenotype showing all recessive traits can be considered as parental. Here, wild-type is (wild) and albino, brown, fuzzy is (albino, brown, fuzzy).
03

Count Recombinant Phenotypes

Recombinant phenotypes show the combinations of alleles that would occur when crossing over happens between linked genes. These are albino (recessive for only this allele), brown (recessive for only this allele), and albino, fuzzy, and brown, fuzzy (which have some combinations of wild-type and recessive alleles).
04

Determine Linkage Status

Based on the phenotypic counts, compare recombinant and parental counts. If parental counts are high relative to recombinants, linkage between genes is likely. A significant difference indicates linkage between albino and the other two genes. Small differences suggest any linkage between albino-brown and albino-fuzzy.
05

Calculate Recombination Frequencies and Map Distances

Recombination frequency for albino-brown = \[ \frac{38 + 42}{1808} \times 100 = 4.4\% \]for brown-fuzzy = \[ \frac{38 + 42}{1790} \times 100 = 4.5\% \]for albino-fuzzy = \[ \frac{3+5}{1760} \times 100 = 0.5\% \]
06

Draw Genetic Map

Using the calculated recombination frequencies, construct a genetic map with albino, brown, and fuzzy genes. Clearly, albino and fuzzy are slightly further away from the main cluster of brown, indicating linkage.
07

Identify Original Genotypes

Using the inferred map distances, establish the original purebred parent phenotypes. One line was homozygous for the wild-type alleles for albino, brown, and fuzzy (\(AL\)\(B\)\(FU\)) and the other was homozygous for the recessive alleles (\(al\)\(b\)\(fu\)).

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 91Ó°ÊÓ!

Key Concepts

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

Recombination Frequency
Recombination frequency is a crucial concept when studying genetic linkage. It signifies the likelihood of crossing over occurring between two linked genes and is calculated as a percentage. This frequency is critical, as it helps us understand how genes are inherited together.

In the case of the rice breeding experiment, recombination frequencies were determined for three gene pairs: albino-brown, brown-fuzzy, and albino-fuzzy. The formula used to calculate recombination frequency is:
  • \[\text{Recombination Frequency} = \left( \frac{\text{Number of Recombinant Offspring}}{\text{Total Number of Offspring}} \right) \times 100\]
This formula helps to quantify how often the linked genes were separated by crossing over during meiosis. For example, a recombination frequency of 4.4% between albino and brown indicates that these genes are fairly close on the chromosome, leading to fewer recombinant offspring.

The recombination frequency is an effective way to predict genetic variation in offspring, especially when examining genetic linkage in breeding experiments. Lower recombination frequencies suggest that genes are more closely linked, while higher frequencies imply the opposite.
Genetic Mapping
Genetic mapping is a significant method used to determine the order of genes and the relative distances between them on a chromosome. This map is established based on recombination frequencies obtained through genetic studies.

In the rice breeding exercise, the genetic map was constructed using the calculated recombination frequencies for albino, brown, and fuzzy genes. These frequencies provide the necessary distances on the map:
  • Albino-brown: 4.4%
  • Brown-fuzzy: 4.5%
  • Albino-fuzzy: 0.5%
These percentages translate into map units or centimorgans (cM), which visually represent the physical distance between genes. The map helps in understanding not only how genes are organized but also how they might interact with each other during inheritance.

Genetic maps are instrumental in identifying gene loci and their role in heredity, and they are particularly valuable in improving plant and animal breeding programs. By understanding these maps, breeders can make more informed decisions when attempting to select for specific traits.
Phenotypic Ratios
Phenotypic ratios are the observable traits expressed in the offspring of genetic crosses. These ratios are essential in identifying whether genes are linked or assort independently.

In this exercise with rice plants, different phenotypes were counted to understand genetic linkage. The progeny included phenotypes such as wild-type, albino, albino-brown, and others. These phenotypic counts provide insight into gene behaviour, specifically if any genes exhibit linkage.

Typical Mendelian inheritance would suggest a certain ratio of phenotypes, for instance, 3:1 or 9:3:3:1. However, the rice experiment yields different phenotypic ratios, indicating that some genes are linked and don't assort independently. For example, the presence of a high number of certain phenotypic classes, such as the parental phenotypes, suggests that genetic linkage is affecting inheritance patterns.

Understanding phenotypic ratios can help students predict the outcome of genetic crosses, identify potential gene linkages, and apply this knowledge to broader genetic studies.

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 a haploid fungus, the genes \(a l-2\) and \(a r g-6\) ar apart on chromosome \(1,\) and the genes \(l y s-5\) \(a l-2+;+m e t-1 \times+\arg -6 ; l y s-5+\) what proportion of progeny would be prototrophic \(++\) \(++?\)

In a tetrad analysis, the linkage arrangement of the \(p\) and \(q\) loci is as follows: Assume that ' in region i, there is no crossover in 88 percent of meioses and there is a single crossover in 12 percent of meioses; in region ii, there is no crossover in 80 percent of meioses and there is a single crossover in 20 percent of meioses; and - there is no interference (in other words, the situation in one region does not affect what is going on in the other region) What proportions of tetrads will be of the following types? (a) \(\mathrm{M}_{\mathrm{I}} \mathrm{M}_{\mathrm{I}}, \mathrm{PD} ;\) (b) \(\mathrm{M}_{\mathrm{I}} \mathrm{M}_{\mathrm{I}}, \mathrm{NPD} ;(\mathrm{c}) \mathrm{M}_{\mathrm{I}} \mathrm{M}_{\mathrm{II}}, \mathrm{T} ;(\mathrm{d}) \mathrm{M}_{\mathrm{II}} \mathrm{M}_{\mathrm{I}}, \mathrm{T} ;\) (e) \(\mathrm{M}_{\mathrm{II}} \mathrm{M}_{\mathrm{II}}, \mathrm{PD} ;(\mathrm{f}) \mathrm{M}_{\mathrm{II}} \mathrm{M}_{\mathrm{II}}, \mathrm{NPD} ;(\mathrm{g}) \mathrm{M}_{\mathrm{II}} \mathrm{M}_{\mathrm{II}}, \mathrm{T}\). (Note: Here the M pattern written first is the one that pertains to the \(p\) locus.) Hint: The easiest way to do this problem is to start by calculating the frequencies of asci with crossovers in both regions, region i, region ii, and neither region. Then determine what \(\mathrm{M}_{\mathrm{I}}\) and \(\mathrm{M}_{\mathrm{II}}\) patterns result.

The \(A\) locus and the \(D\) locus are so tightly linked that no recombination is ever observed between them. If \(A d /\) \(A d\) is crossed with \(a D / a D\) and the \(F_{1}\) is intercrossed, what phenotypes will be seen in the \(\mathrm{F}_{2}\) and in what proportions?

The Neurospora cross al-2 \(^{+} \times\) al-2 is made. A linear tetrad analysis reveals that the second-division segregation frequency is 8 percent. a. Draw two examples of second-division segregation patterns in this cross. b. What can be calculated by using the 8 percent value?

The five sets of data given in the following table represent the results of testcrosses using parents with the same alleles but in different combinations. Determine the order of genes by inspection-that is, without calculating recombination values. Recessive phenotypes are symbolized by lowercase letters and dominant phenotypes by pluses.

See all solutions

Recommended explanations on Biology Textbooks

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.