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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)?

Short Answer

Expert verified
7.68% of the progeny will have the desired phenotype if all conditions hold.

Step by step solution

01

Parent Chromosome Diagrams

Begin by outlining the chromosomes from both parents. The chromosome of parent 1 is \( A A L L d d \), meaning it holds alleles for anthocyanin and long tassels and is homozygous for dwarfism recessive. Draw this as a single chromatid with these genes. Parent 2 is \( a a U D D \) which, as a pure line, presents the dominant phenotype for the dwarf plant and another unknown phenotype at \( U \). The fully recessive line is \( a a u d d \).
02

Diagram of the Hybrid

Cross the alleles of the first two parents to form the first generation hybrid genotype \( A a L l D d \). Diagram this hybrid showing the hybrid chromosomes, composed of alleles from the parents: one chromosome coming from \( A A L L d d \) and another from \( a a U D D \).
03

Diagram of the Tester

The tester, or the fully recessive parent, is summarized as \( a a u d d \). This chromosome should be straightforward with the combination being homozygous recessive for all traits. Draw this as a dashed line to show all relevant genes are recessive alleles.
04

Crossover Event Diagrams

The desired phenotype \( Aa Ll Dd \) can be produced through specific crossover events between \( ALd \) and \( aUD \). Draw the recombination process, focusing on crossing over between linked genes. Show double crossovers: one between \( A/a \) and \( L/l \), and one between \( L/l \) and \( D/d \), as these knockout the neighboring alleles to produce the required combination.
05

Calculate Crossover Probability

Calculate the double crossover probability given by the product of the individual crossover probabilities: \( 0.16 \times 0.24 \), which is the distance in map units between the genes expressed as probabilities.
06

Testcross Phenotype Calculation

Multiply the probability of a double crossover by 2 to consider the two potential recombinant products. Realize the desired phenotype is a recombinant type. Therefore, the final progeny with the desired \( AaLlDd \) phenotype is \( 2 \times 0.16 \times 0.24 = 0.0768 \) or 7.68%.
07

Assumptions Analysis

Acknowledge any assumptions made, chief among them an assumption of complete dominance of the dominant alleles over the recessive alleles, large enough progeny to express all predicted phenotypes, and that crossover frequencies are similar across the population and can be estimated from map units.

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Key Concepts

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

Linkage and Recombination
In genetics, linkage refers to the tendency of genes that are located close to each other on a chromosome to be inherited together during meiosis. This happens because they are physically connected and less likely to be separated by recombination. Recombination, on the other hand, involves the exchange of genetic material between homologous chromosomes, resulting in mixtures of parental characteristics in offspring.

When genes are linked, they do not assort independently, leading to non-Mendelian inheritance ratios. In our corn example, the genes for anthocyanin (A), long tassels (L), and dwarf plant (D) are linked on the same chromosome.
  • This means that these genes usually travel together during the formation of gametes.
  • The linkage is broken when crossover events occur between these genes, allowing new combinations to be formed.
Understanding linkage and recombination is crucial for predicting genetic outcomes and breeding strategies at the geneticist's disposal.
Genotype and Phenotype
Genotype refers to the genetic makeup of an organism. It encompasses all the alleles that an organism possesses, whether they are expressed visibly or not. Phenotype, on the other hand, is the observable characteristics or traits of an organism, which result from the interaction of its genotype with the environment.

In our exercise, when the geneticist combined two corn plants, the goal was to obtain plants with the dominant phenotypes for certain traits. Although they were working with specific genotypes, the geneticist wanted a phenotype characterized by anthocyanin, long tassels, and dwarf stature.
  • The genotype Aa Ll Dd results in the desired phenotype, thanks to the presence of dominant alleles "A," "L," and "D."
  • While Aa indicates heterozygosity, as does Ll and Dd, the phenotype exhibits dominant traits because dominant alleles mask the effect of recessive ones.
Understanding the relationship between genotype and phenotype helps in forecasting which traits will appear in offspring and planning breeding programs effectively.
Crossover Probability
Crossover probability refers to the likelihood of genetic recombination happening between genes during meiosis. This affects the genetic diversity of the offspring as different combinations of alleles can be formed through these crossovers.

Mapping genes based on crossover probability is an essential part of genetic analysis. Map units (mu) are used to provide a measure of the likelihood that a crossover will occur between two genes. In our example, there are 16 map units between "A/a" and "L/l" and 24 map units between "L/l" and "D/d".
  • The probability of a single crossover event occurring in a given region is directly related to the number of map units between the genes.
  • The probability of a double crossover, which is necessary to obtain the desired genotype in our exercise, is the product of the individual crossover probabilities.
For this specific geneticist's goal, calculating crossover probability allowed them to anticipate that approximately 7.68% of the progeny would exhibit the desired phenotype, a valuable insight for planning genetic experiments.

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Most popular questions from this chapter

In corn, a triple heterozygote was obtained carrying the mutant alleles \(s\) (shrunken), \(w\) (white aleurone), and \(y\) (waxy endosperm), all paired with their normal wild-type alleles. This triple heterozygote was testcrossed, and the progeny contained 116 shrunken, white; 4 fully wild type; 2538 shrunken; 601 shrunken, waxy; 626 white; 2708 white, waxy; 2 shrunken, white, waxy; and 113 waxy. a. Determine if any of these three loci are linked and, if so, show map distances. b. Show the allele arrangement on the chromosomes of the triple heterozygote used in the testcross. c. Calculate interference, if appropriate.

The cross \(E / E \cdot F / F \times e / e \cdot f / f\) is made, and the \(F_{1}\) is then backcrossed with the recessive parent. The prog. eny genotypes are inferred from the phenotypes. The progeny genotypes, written as the gametic contributions of the heterozygous parent, are in the following proportions: \(E \cdot F \quad \frac{2}{6}\) \(E \cdot f \quad \frac{1}{6}\) \(e \cdot F \quad \frac{1}{6}\) \(e \cdot f \quad \frac{2}{6}\) Explain these results.

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?

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. $$\begin{array}{ccccccc} \mathbf{1} & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline A \cdot \text {arg} & A \cdot+ & A \cdot \arg & A \cdot \text {arg} & A \cdot \text {arg} & A \cdot+ & A \cdot+ \\ A \cdot \text {arg} & A \cdot+ & A \cdot+ & a \cdot \arg & a \cdot+ & a \cdot \text {arg} & a \cdot \arg \\ a \cdot+ & a \cdot \arg & a \cdot \arg & A \cdot+ & A \cdot \arg & A \cdot+ & A \cdot \arg \\ a \cdot+ & \frac{a \cdot \arg }{127} & \frac{a \cdot+}{100} & \frac{a \cdot+}{36} & \frac{a \cdot+}{2} & \frac{a \cdot \arg }{4} & \frac{a \cdot+}{6} \end{array}$$

An individual heterozygous for four genes, \(A / a \cdot B / b \cdot C / c \cdot D / d\) is testcrossed with \(a / a \cdot b / b \cdot c / c \cdot d / d,\) and 1000 progeny are classified by the gametic contribution of the heterozygous parent as follows: $$\begin{array}{cc} a \cdot B \cdot C \cdot D & 42 \\ A \cdot b \cdot c \cdot d & 43 \\ A \cdot B \cdot C \cdot d & 140 \\ a \cdot b \cdot c \cdot D & 145 \\ a \cdot B \cdot c \cdot D & 6 \\ A \cdot b \cdot C \cdot d & 9 \\ A \cdot B \cdot c \cdot d & 305 \\ a \cdot b \cdot C \cdot D & 310 \end{array}$$ a. Which genes are linked? b. If two pure-breeding lines had been crossed to produce the heterozygous individual, what would their genotypes have been? c. Draw a linkage map of the linked genes, showing the order and the distances in map units. d. Calculate an interference value, if appropriate.

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