/*! 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 53 From parents of genotypes \(A / ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

From parents of genotypes \(A / A \cdot B / B\) and \(a / a \cdot b / b, a\) dihybrid was produced. In a testcross of the dihybrid, the following seven progeny were obtained: \\[\begin{array}{c}A / a \cdot B / b, a / a \cdot b / b, A / a \cdot B / b, A / a \cdot b / b \\ a / a \cdot b / b, A / a \cdot B / b, \text { and } a / a \cdot B / b\end{array}\\] Do these results provide convincing evidence of linkage?

Short Answer

Expert verified
The small sample size makes it difficult to conclude linkage definitively.

Step by step solution

01

Understanding Parental Genotype

The parental genotypes are \(A/A \cdot B/B\) and \(a/a \cdot b/b\). This means that one parent is homozygous dominant for both genes, and the other is homozygous recessive for both genes.
02

Determine Dihybrid Genotype

A dihybrid resulting from these parents will have the genotype \(A/a \cdot B/b\), as it receives one dominant and one recessive allele for both genes from its parents.
03

Define Testcross

In a testcross, we cross the dihybrid \(A/a \cdot B/b\) with a homozygous recessive individual, \(a/a \cdot b/b\). This helps to reveal the recombinant and parental genotypes in the offspring.
04

Calculate Expected Ratios for Independent Assortment

If the two genes were assorting independently, we would expect a 1:1:1:1 ratio of the possible genotypes: \(A/a \cdot B/b\), \(A/a \cdot b/b\), \(a/a \cdot B/b\), \(a/a \cdot b/b\).
05

Analyze the Progeny

The offspring given are: \(A/a \cdot B/b, a/a \cdot b/b, A/a \cdot B/b, A/a \cdot b/b, a/a \cdot b/b, A/a \cdot B/b, a/a \cdot B/b\). Count the occurrences: \(A/a \cdot B/b\) occurs 3 times, \(a/a \cdot b/b\) occurs twice, \(A/a \cdot b/b\) occurs once, and \(a/a \cdot B/b\) occurs once.
06

Conclusion on Linkage

The results do not perfectly fit the 1:1:1:1 ratio expected from independent assortment, but with only 7 progeny, the sample size is too small to make a definitive conclusion about linkage. The observed numbers suggest possible linkage, but more data is needed for a convincing argument.

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.

Dihybrid Cross
A dihybrid cross involves organisms that are heterozygous for two different traits. In our exercise, we have parents with genotypes \( A/A \cdot B/B \) and \( a/a \cdot b/b \). The offspring of this cross, or the dihybrid, will all have the genotype \( A/a \cdot B/b \). This is because each parent contributes one allele for each of the two genes. The dominant and recessive alleles combine to produce heterozygous offspring.
The importance of a dihybrid cross lies in studying how traits are inherited when more than one gene is involved. It's a tool to examine how combinations arise in the progeny, and when combined with a testcross, as in our exercise, it helps uncover genetic linkage.
Testcross
A testcross is an experiment that can reveal the genotype of an organism with a dominant phenotype. In this context, the dihybrid \( A/a \cdot B/b \) is crossed with a homozygous recessive individual \( a/a \cdot b/b \). Why use a homozygous recessive? Because its known genotype allows any dominant characteristics from the heterozygous parent to be clearly expressed in the progeny.
In our exercise, the testcross aids in uncovering the combinations of alleles that result from the dihybrid parent. By evaluating the offspring's genotypes, we can infer the patterns of linkage — or lack thereof — between the two traits being studied. This makes the testcross a critical component in genetic experiments aimed at mapping genes.
Independent Assortment
Independent assortment describes how genes located on different chromosomes are inherited independently of each other during meiosis. The result, with no linkage, is the classic 9:3:3:1 phenotypic ratio in a dihybrid cross or a 1:1:1:1 phenotypic ratio in a testcross like ours.
In the exercise, under independent assortment, the four possible genotypic combinations from the dihybrid testcross should appear in equal ratios. Our observed progeny numbers slightly deviate from this (3:1:1:2) which suggests other factors might be influencing the inheritance pattern, like linkage. Yet, with limited progeny, as in many real-world scenarios, these deviations could also just be due to chance.
Homozygous and Heterozygous Genotypes
To fully understand genetic crosses, it's crucial to grasp the meanings of homozygous and heterozygous genotypes. Homozygous individuals have two identical alleles for a gene, either dominant \( (e.g., A/A) \) or recessive \( (e.g., a/a) \). Heterozygous individuals have one dominant and one recessive allele \( (e.g., A/a) \).
In our exercise, the parents are homozygous for the genes being studied, providing clear dominance or complete recessiveness in traits. The resulting dihybrid, and hence the testcross progeny, are mixtures of these to show the allele combinations inherited from parents. Distinguishing between these genotypes is fundamental to predict genetic outcomes and to test for phenomena like linkage or independent assortment.

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

A Neurospora cross was made between a strain that carried the mating-type allele \(A\) and the mutant allele arg-1 and another strain that carried the matingtype allele \(a\) and the wild-type allele for \(\arg -1(+)\) Four hundred linear octads were isolated, and they fell into the seven classes given in the table below. (For simplicity, they are shown as tetrads.) a. Deduce the linkage arrangement of the mating-type locus and the arg-1 locus. Include the centromere or centromeres on any map that you draw. Label all intervals in map units. b. Diagram the meiotic divisions that led to class 6 Label clearly. $$\begin{array}{ccccccc} \mathbf{1} & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline A \cdot \text {arg} & A \cdot+ & A \cdot \arg & A \cdot \text {arg} & A \cdot \text {arg} & A \cdot+ & A \cdot+ \\ A \cdot \text {arg} & A \cdot+ & A \cdot+ & a \cdot \arg & a \cdot+ & a \cdot \text {arg} & a \cdot \arg \\ a \cdot+ & a \cdot \arg & a \cdot \arg & A \cdot+ & A \cdot \arg & A \cdot+ & A \cdot \arg \\ a \cdot+ & \frac{a \cdot \arg }{127} & \frac{a \cdot+}{100} & \frac{a \cdot+}{36} & \frac{a \cdot+}{2} & \frac{a \cdot \arg }{4} & \frac{a \cdot+}{6} \end{array}$$

In a haploid fungus, the genes al-2 and arg-6 are 30 map units apart on chromosome \(1,\) and the genes lys- 5 and met-1 are 20 map units apart on chromosome 6. In a cross \\[a l-2+;+m e t-1 \times+\arg -6 ; 1 y s-5+\\] what proportion of progeny would be prototrophic \\[++;++?\\]

A corn geneticist wants to obtain a corn plant that has the three dominant phenotypes: anthocyanin (A), long tassels (L), and dwarf plant (D). In her collection of pure lines, the only lines that bear these alleles are \(A A L L d d\) and aa \(U D D\). She also has the fully recessive line aa \(u\) dd. She decides to intercross the first two and testcross the resulting hybrid to obtain in the progeny a plant of the desired phenotype (which would have to be Aa Ll \(D d\) in this case). She knows that the three genes are linked in the order written and that the distance between the \(A / a\) and the \(L / l\) loci is 16 map units and that the distance between the \(L / l\) and the \(D / d\) loci is 24 map units. a. Draw a diagram of the chromosomes of the parents, the hybrid, and the tester. b. Draw a diagram of the crossover(s) necessary to produce the desired genotype. c. What percentage of the testcross progeny will be of the phenotype that she needs? d. What assumptions did you make (if any)?

A female animal with genotype \(A / a \cdot B / b\) is crossed with a double- recessive male \((a / a \cdot b / b) .\) Their progeny include 442 A/a \(\cdot B / b, 458 a / a \cdot b / b, 46 A / a \cdot b / b,\) and 54 \(a / a \cdot B / b .\) Explain these results.

Chromosome 3 of corn carries three loci ( \(b\) for plantcolor booster, \(v\) for virescent, and \(\lg\) for liguleless). \(A\) testcross of triple recessives with \(\mathrm{F}_{1}\) plants heterozygous for the three genes yields progeny having the following genotypes: \(305+v\) lg, \(275 b++128 b+l g\) \(112+v+, 74++\lg , 66 b v+, 22+++,\) and 18 \(b v\) lg. Give the gene sequence on the chromosome, the map distances between genes, and the coefficient of coincidence.

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.