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Chapter 4: Problem 12

A plant of genotype $$\begin{array}{cc} A & B \\\\\hline \hline a & b\end{array}$$ is testcrossed with $$\begin{array}{cc}a & b \\\\\hline \hline a & b\end{array}$$ If the two loci are 10 m.u. apart, what proportion of progeny will be \(A B / a b ?\)

Short Answer

Expert verified
The proportion of progeny with genotype \( A B / a b \) is 45%.

Step by step solution

01

Understanding the Problem

We have a plant with heterozygous genotype \( A B / a b \) which is testcrossed with a homozygous recessive plant \( a b / a b \). We need to find out the proportion of progeny showing the genotype \( A B / a b \), knowing that the loci are 10 map units apart, indicating 10% recombination rate.
02

Define Parental and Recombinant Types

The parental genotypes from the testcross will be \( A B / a b \) and \( a b / a b \). Recombinant genotypes will be \( A b / a b \) and \( a B / a b \). Since the loci are 10 map units apart, 10% of offspring will be recombinants, and 90% will be parental.
03

Calculate Recombinant and Parental Proportions

Since 10% of the offspring are expected to be the result of recombination, 5% will be \( A b / a b \) and 5% will be \( a B / a b \). The remaining 90% will be split between the parental types \( A B / a b \) and \( a b / a b \).
04

Calculate Proportion of \( A B / a b \)

As 90% of the offspring are parental, and the parental genotypes are equally probable, the proportion of \( A B / a b \) will be half of 90%, which is 45%.

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Key Concepts

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

Map Units
In genetics, map units (m.u.) are a way to measure the distance between two genes on a chromosome. The distance is related to the frequency with which recombination occurs between them. One map unit is equivalent to a 1% recombination rate.

This means if you have two genes that are 10 map units apart, like in our exercise, there is a 10% chance of recombination between these two genes during the formation of egg or sperm cells.

  • Map units help us understand the physical proximity of genes on a chromosome.
  • A higher number of map units (e.g., 50 m.u.) suggests a greater distance, and thus, a higher chance of recombination.
  • This concept aids in predicting genetic outcomes in offspring.
By using map units, scientists can construct linkage maps that visually represent the relative positions of genes on a chromosome. This is crucial for understanding genetic linkage and for plant and animal breeding.
Testcross
A testcross is a breeding experiment used to determine the genotype of an individual exhibiting a dominant phenotype.

In our exercise, the testcross involves a heterozygous plant (one dominant allele for each gene) crossed with a homozygous recessive plant (both alleles recessive). The main purpose here is to reveal whether the dominant phenotype is linked to a homozygous or heterozygous genotype.
  • The offspring produced help show whether certain genes are linked (inherited together) or show some degree of recombination.
  • If recombination occurs, it will result in new combinations of traits in the progeny.
  • This method is powerful in exploring genetic recombination and understanding gene linkage.
Through testcrossing, we obtain actionable data on how linked or independent various genes are, helping to map comprehensive genetic pathways.
Progeny Genotype Proportions
Understanding progeny genotype proportions is essential in predicting genetic outcomes from crosses.

In this exercise, the possible progeny types arise from the combination of alleles during meiosis. The distance between loci (measured in map units) predicts the proportion of these genotypes in the offspring.

Since the loci are 10 map units apart, 10% of the progeny will be recombinant. Meaning they will exhibit new allele combinations such as:
  • 5% as \(A b / a b\)
  • 5% as \(a B / a b\)
The rest, 90% of progeny, will be parental types, maintaining the original combinations from the parents. Parental types here will split evenly between \(A B / a b\) and \(a b / a b\), resulting in 45% each.
  • This method of prediction provides insight into genetic variability.
  • Subsequently allowing geneticists to predict traits in future generations with greater certainty.
Comprehending progeny genotype proportions is, therefore, pivotal for controlling and enhancing the genetic characteristics of plants and animals.

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Most popular questions from this chapter

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?

The recessive alleles \(k\) (kidney-shaped eyes instead of wild-type round), \(c\) (cardinal-colored eyes instead of wildtype red), and \(e\) (ebony body instead of wild-type gray) identify three genes on chromosome 3 of Drosophila. Females with kidney-shaped, cardinal-colored eyes were mated with ebony males. The \(\mathrm{F}_{1}\) was wild type. When \(\mathrm{F}_{1}\) females were testcrossed with \(k k\) cc ee males, the following progeny phenotypes were obtained: \(\begin{array}{cccr}k & c & e & 3 \\ k & c & \+ & 876 \\ k & \+ & e & 67 \\\ k & \+ & \+ & 49 \\ + & c & e & 44 \\ + & c & \+ & 58 \\ + & \+ & e & 899 \\\ + & \+ & \+ & 4 \\ \text { Total } & & & 2000\end{array}\) a. Determine the order of the genes and the map distances between them. b. Draw the chromosomes of the parents and the \(\mathrm{F}_{1}\). c. Calculate interference and say what you think of its significance.

For a certain chromosomal region, the mean number of crossovers at meiosis is calculated to be two per meiosis. In that region, what proportion of meioses are predicted to have (a) no crossovers? (b) one crossover? (c) two crossovers?

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 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 \(L h / l h ?\)
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