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Chapter 4: Problem 15

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.

Short Answer

Expert verified
The ratios indicate linkage between genes \(E\) and \(F\) with some recombination, as seen from the higher proportion of parental gametes.

Step by step solution

01

Understanding the initial cross

The initial cross is between two parents: one homozygous dominant for both traits \(E / E \cdot F / F\) and one homozygous recessive \(e / e \cdot f / f\). This produces \(F_{1}\) offspring that are all heterozygous \(E / e \cdot F / f\).
02

Analyzing the backcross

The \(F_1\) generation \(E / e \cdot F / f\) is then backcrossed with the homozygous recessive parent \(e / e \cdot f / f\). This means the \(F_1\)'s contributions can be in the form of its gametes: \(E \cdot F, E \cdot f, e \cdot F,\) and \(e \cdot f\).
03

Determining progeny ratios

The progeny exhibit the following genotypes from the \(F_1\)'s gametic combinations: \(E \cdot F\) (2/6), \(E \cdot f\) (1/6), \(e \cdot F\) (1/6), and \(e \cdot f\) (2/6). These ratios indicate the likelihood of each gametic type being passed on.
04

Explaining the results using linkage and recombination

Given the disproportional ratios, the genes \(E/e\) and \(F/f\) are likely linked, showing recombination. The parental-type gametes \(E \cdot F\) and \(e \cdot f\) are more common (2/6 each), while the recombinant-type gametes \(E \cdot f\) and \(e \cdot F\) are less frequent (1/6 each). This implies linkage with partial recombination occurring.

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Key Concepts

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

Backcross
A backcross is a common method in genetics and is conducted to understand how traits are inherited in offspring. It involves crossing a hybrid organism, which is heterozygous for certain traits, back to one of its parents, often the recessive one. This process helps to reveal the genetic makeup of the organism undergoing the backcross.

In the context of our exercise, the initial hybrid, or \( F_1 \) generation, with the genotype \( E / e \cdot F / f \), was crossed back with the homozygous recessive parent \( e / e \cdot f / f \).
This allows researchers to analyze how genes from the \( F_1 \) generation are passed on to the next, unveiling information about gene linkage and recombination events.
  • The hybrid organism is heterozygous, carrying two different alleles for the same trait.
  • Backcrossing helps determine if certain traits are linked or unlinked.
By looking at the distribution of traits in the progeny, scientists can infer potential linkages between specific genes.
Gametes
Gametes are the reproductive cells used in sexual reproduction to create new organisms. They carry a single set of chromosomes and are crucial for transmitting genetic information from parents to offspring. In this exercise, we aim to understand how different gametes contribute to the genetic makeup of the progeny.

Gametes fuse during fertilization to form a new organism with a complete set of chromosomes. Each gamete carries one allele for each gene, contributing to the offspring's genetic diversity.
  • The types of gametes released by the \( F_1 \) generation in the exercise are: \( E \cdot F \), \( E \cdot f \), \( e \cdot F \), and \( e \cdot f \).
  • These combinations represent different possible allele configurations that can be passed to the next generation.
Understanding the proportions in which these gametes are produced can offer insights into genetic linkage and inheritance patterns.
Recombination
Recombination is a genetic process occurring during the formation of gametes, where segments of DNA are shuffled to create new allele combinations. This can greatly increase genetic diversity among offspring. It involves the physical exchange of chromosome segments, leading to new combinations of alleles in gametes.

In our exercise, recombination plays a key role in generating the observed gametic proportions. The expected 1:1:1:1 ratio in a situation without linkage is skewed, pointing to recombination action.
  • Parental-type gametes often appear more frequently, indicating limited recombination.
  • Recombination can sometimes be measured to map gene locations on chromosomes.
This particular exercise demonstrates that while recombination is occurring, it is not complete, suggesting genetic linkage between the \( E/e \) and \( F/f \) genes.
Heterozygous
Heterozygous organisms have two different alleles for a particular trait. This is central to our exercise, where the \( F_1 \) generation organisms are heterozygous for both traits involved: \( E/e \) and \( F/f \).

These heterozygous organisms provide a rich source of genetic variation, as they can produce a variety of gametes carrying different combinations of alleles. This variation is crucial for studying how traits are inherited across generations.
  • Heterozygosity allows for the expression of both dominant and recessive traits.
  • It is essential for understanding phenomena like incomplete dominance and codominance.
In the context of this experiment, the heterozygous nature of \( F_1 \) individuals contributes significantly to the diverse genetic outcomes in the progeny when backcrossed with a homozygous recessive parent.

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Most popular questions from this chapter

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.

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 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 \(\mathrm{F}_{1}\) is testcrossed, what proportion of progeny will be ww ee ff? b. If the \(\mathrm{F}_{1}\) is selfed, what proportion of progeny will be ww ee ff?

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) \(\mathrm{M}_{\mathrm{f}} \mathrm{M}_{\mathrm{I}}, \mathrm{PD}\) \((\mathbf{b}) \mathrm{M}_{1} \mathrm{M}_{\mathrm{l}}, \mathrm{NPD}\) \((\mathbf{c}) M_{1} M_{11}, T ;\) (d) \(\mathrm{M}_{\mathrm{D}} \mathrm{M}_{\mathrm{l}}, \mathrm{T}\) \((\mathbf{e}) M_{1} M_{\Perp}, P D\) \((\mathbf{f}) M_{11} M_{11}, N P D ;\) (g) \(\mathrm{M}_{\mathrm{II}} \mathrm{M}_{\mathrm{U}}\), 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 \(\mathrm{M}_{1}\) and \(\mathrm{M}_{\mathrm{II}}\) patterns result.

The Neurospora cross al- \(2^{+} \times a l-2\) 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?
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