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Chapter 4: Problem 18

If \(A / A \cdot B / B\) is crossed with \(a / a \cdot b / b\) and the \(F_{1}\) is testcrossed, what percentage of the testcross progeny will be \(a / a \cdot b / b\) if the two genes are (a) unlinked; (b) completely linked (no crossing over at all); (c) \(10 \mathrm{m}\).u. apart; (d) \(24 \mathrm{m} .\) u. apart?

Short Answer

Expert verified
(a) 25% (b) 50% (c) 45% (d) 38%

Step by step solution

01

Understanding the Genotypes

We begin with the genotypes of the parental generations:- Parental 1: \( A / A \cdot B / B \)- Parental 2: \( a / a \cdot b / b \)The \( F_1 \) generation obtained from crossing these parents will result in all \( A / a \cdot B / b \).
02

Testcross Set-up

The \( F_1 \) genotype \( A/a \cdot B/b \) is then testcrossed with the homozygous recessive genotype \( a/a \cdot b/b \). The testcross involves crossing \( A/a \cdot B/b \) with \( a/a \cdot b/b \).
03

Case a) Unlinked Genes

When the genes are unlinked, they assort independently according to Mendel's second law. Therefore, the testcross progeny will be:- 25% \( A/a \cdot B/b \)- 25% \( A/a \cdot b/b \)- 25% \( a/a \cdot B/b \)- 25% \( a/a \cdot b/b \)Thus, the percentage of \( a/a \cdot b/b \) is 25%.
04

Case b) Completely Linked Genes

If the genes are completely linked with no crossing over, the parental alleles are inherited together. Thus, all testcross progeny will inherit:- 0% recombinants, meaning 50% \( A/a \cdot B/b \) and 50% \( a/a \cdot b/b \)Therefore, the percentage of \( a/a \cdot b/b \) is 50%.
05

Case c) 10 m.u. Apart

The genetic distance is 10 map units, indicating 10% recombinant offspring. Therefore:- 90% parental types (45% each for \( A/a \cdot B/b \) and \( a/a \cdot b/b \))- 10% recombinant types (5% each for \( A/a \cdot b/b \) and \( a/a \cdot B/b \))The percentage of \( a/a \cdot b/b \) is 45%.
06

Case d) 24 m.u. Apart

The genetic distance is 24 map units, indicating 24% recombinant offspring. Therefore:- 76% parental types (38% each for \( A/a \cdot B/b \) and \( a/a \cdot b/b \))- 24% recombinant types (12% each for \( A/a \cdot b/b \) and \( a/a \cdot B/b \))The percentage of \( a/a \cdot b/b \) is 38%.

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Key Concepts

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

Testcross
In genetics, a testcross is a powerful method to determine the genetic makeup, or genotype, of an organism. It involves crossing an organism possessing a dominant phenotype (but unknown genotype) with a homozygous recessive organism.
This helps identify whether the organism with the dominant trait is homozygous dominant or heterozygous.
  • A homozygous dominant testcross results in offspring that all display the dominant phenotype.
  • A heterozygous testcross will yield a mix of the dominant and recessive phenotypes.
In our exercise, the process involves crossing the first-generation offspring (noted as the \( F_1 \) generation) with the homozygous recessive genotype \( a / a \cdot b / b \).
This facilitates the observation of how alleles segregate and recombine, offering insights into genetic linkage and assortment patterns.
Recombination
Recombination refers to the process by which chromosomes exchange pieces of their genetic material during meiosis. This exchange is vital in contributing to genetic diversity, as it results in new combinations of alleles.
During a testcross, the degree of recombination between genes can significantly affect the outcome of offspring genotypes.
  • If genes are unlinked, or far apart on a chromosome, they assort independently. This follows Mendel's second law of independent assortment. Each possible allele combination occurs in equal proportions.
  • Complete linkage occurs when genes are so close together on a chromosome that they are usually inherited together. Little to no recombination happens between them.
  • Partial linkage indicates some degree of recombination between genes, quantified by map units (m.u.) in gene mapping.
In the exercise, recombination determines what percentage of offspring exhibit a specific genotype, such as \( a/a \cdot b/b \). The genetic distance, given in map units, indicates likelihood and frequency of recombination events.
Gene Mapping
Gene mapping is a technique used to determine the position of genes on the chromosomes and their genetic distance from each other.
This method offers essential insights into how frequently genes recombine during meiosis. The unit of measure for the distance between genes is referred to as a "map unit" (m.u.).
  • 1 map unit equals a 1% recombination frequency, meaning a 1% chance that a crossover event will occur between two genes for every 100 offspring.
  • Genes that are 10 m.u. apart have a 10% recombination rate.
  • The further apart genes are, up to 50 map units, the higher the recombination frequency.
In our exercise scenarios, understanding gene mapping helps predict the percentage of offspring with specific genotypes, depending on the genetic distance between the genes \( A/a \) and \( B/b \). This information is critical for deciphering genetic linkage patterns and for practical applications in breeding and genetics research.

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Most popular questions from this chapter

In the plant Arabidopsis, the loci for pod length (L, long; 1, short) and fruit hairs ( \(H,\) hairy; \(h,\) smooth) are linked 16 m.u. apart on the same chromosome. The following crosses were made: (i) \(L H / L H \times l h / l h \rightarrow F_{1}\) (ii) \(L h / L h \times l H / l H \rightarrow F_{1}\) If the \(\mathrm{F}_{1}\) 's from cross i and cross ii are crossed, a. what proportion of the progeny are expected to be \(l h / l h ?\) b. what proportion of the progeny are expected to be \(L h / l h ?\)

In corn, the cross \(W W\) ee \(F F \times w w E E f f\) is made. The three loci are linked as follows Assume no interference. a. If the \(F_{1}\) is testcrossed, what proportion of progeny will be ww ee ff? b. If the \(F_{1}\) is selfed, what proportion of progeny will be ww ee ff?

In a haploid organism, the \(C\) and \(D\) loci are 8 m.u. apart. From a cross \(C d \times c D\), give the proportion of each of the following progeny classes: (a) \(C D ;\) (b) \(c d ;\) (c) \(C d ;\) (d) all recombinants combined.

A strain of Neurospora with the genotype \(H \cdot I\) is crossed with a strain with the genotype \(h \cdot i\). Half the progeny are \(H \cdot I,\) and the other half are \(h \cdot i .\) Explain how this outcome is possible.

R. A. Emerson crossed two different pure-breeding lines of corn and obtained a phenotypically wild-type \(\mathrm{F}_{1}\) that was heterozygous for three alleles that determine recessive phenotypes: an determines anther; br, brachytic; and \(f,\) fine. He testcrossed the \(\mathrm{F}_{1}\) with a tester that was homozygous recessive for the three genes and obtained these progeny phenotypes: 355 anther; 339 brachytic, fine; 88 completely wild type; 55 anther, brachytic, fine; 21 fine; 17 anther, brachytic; 2 brachytic; 2 anther, fine. a. What were the genotypes of the parental lines? b. Draw a linkage map for the three genes (include map distances). c. Calculate the interference value.
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