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Chapter 4: Problem 13

The \(A\) locus and the \(D\) locus are so tightly linked that no recombination is ever observed between them. If \(A d / A d\) is crossed with \(a D / a D\) and the \(F_{1}\) is intercrossed, what phenotypes will be seen in the \(\mathrm{F}_{2}\) and in what proportions?

Short Answer

Expert verified
The F2 phenotypes will include: 1 \(A d / A d\), 2 \(A d / a D\), and 1 \(a D / a D\), in a 1:2:1 ratio.

Step by step solution

01

Understanding of Genetic Linkage

Since the \(A\) and \(D\) loci are tightly linked with no recombination observed, these two loci will travel together during gamete formation. This means that alleles from each parent will remain together as they are inherited.
02

Initial Genotypes of Parent Generation

The first parent has the genotype \(A d / A d\), and the second parent has the genotype \(a D / a D\). These genotypes imply that each parent can only produce one type of gamete: \(A d\) from the first parent and \(a D\) from the second parent.
03

F1 Generation and Genotype

When the \(A d\) and \(a D\) gametes combine, the resulting \(F_1\) generation's genotype is \(A d / a D\). All \(F_1\) individuals will have this genotype since there is no recombination.
04

Formation of F2 Generation

The \(F_1\) generation \(Ad / aD\) is intercrossed. During gamete formation, each \(F_1\) will produce two types of gametes: \(Ad\) and \(aD\). Thus, the possible combinations for the \(F_2\) are generated from the union of these gametes.
05

Determine F2 Phenotypes and Proportions

When the different gametes from \(F_1\) individuals combine, the possible \(F_2\) genotypes and resulting phenotypes are:1. \(A d / A d\) from \(Ad\) and \(Ad\) - phenotype: both alleles from parent 1.2. \(A d / a D\) from \(Ad\) and \(aD\) - phenotype: phenotype of \(F_1\) individuals.3. \(a D / a D\) from \(aD\) and \(aD\) - phenotype: both alleles from parent 2.4. \(a D / A d\) from \(aD\) and \(Ad\) - phenotype: same as \(A d / a D\).The expected phenotypic ratios are 1:2:1.

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Key Concepts

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

Phenotypic Ratios
When studying genetic linkage and inheritance, understanding phenotypic ratios is crucial. In genetics, a phenotypic ratio refers to the relative number of offspring displaying particular traits as a result of genetic crossing. In this context, we are observing the phenotypes resulting from crossing linked genes.

In the exercise, the phenotypic ratio of the \( F_2 \) generation was determined to be 1:2:1. This means:
  • One part of the offspring will show the phenotype inherited from the first parent \( (Ad/Ad) \).
  • Two parts will display the mixed phenotype of the \( F_1 \) generation \( (Ad/aD) \) or \( (aD/Ad) \) which results in the same phenotype.
  • One part will show the phenotype inherited from the second parent \( (aD/aD) \).
This understanding helps predict how traits are inherited and expressed in the next generation. By showing a 1:2:1 phenotypic ratio, it underscores the Mendelian genetics principle while emphasizing the effect of genetic linkage.
Gamete Formation
Gamete formation is a key part of the inheritance process and includes the creation of sex cells (sperm and egg cells) during meiosis. During this process, alleles (different forms of a gene) are distributed to gametes. In tightly linked loci, like the ones in our example, the alleles are inherited together without recombination.

In the given scenario, the parental genotypes \( Ad/Ad \) and \( aD/aD \) illustrate genetic linkage, where no crossover occurs between the linked alleles of \( A \) and \( D \). Thus, only specific gametes can form:
  • The first parent, \( Ad/Ad \), forms gametes with the alleles \( Ad \).
  • The second parent, \( aD/aD \), forms gametes with \( aD \).
As a result, when these gametes combine in the \( F_1 \) generation, each offspring receives the \( Ad \) allele from one parent and the \( aD \) allele from the other, forming a specific combination without variation from recombination.
F1 and F2 Generations
In genetics, understanding the progression from the \( F_1 \) to the \( F_2 \) generations is essential for illustrating inheritance patterns. The \( F_1 \) generation represents the first generation of offspring resulting from the parental cross. In our example, crossing the genotypes \( Ad/Ad \) and \( aD/aD \) leads to an \( F_1 \) generation with all individuals having the genotype \( Ad/aD \). This uniformity in the \( F_1 \) generation鈥檚 genotype is due to the absence of recombination, leading to the inheritance of complete sets of linked alleles. The \( F_2 \) generation is produced when \( F_1 \) individuals intermate. Even with linked loci, gametes have full genes, causing new combinations within the \( F_2 \) generation, generating diverse phenotypes based on the recombined genotypes. In this specific scenario:
  • \( Ad/Ad \) individuals are formed from \( Ad \) gametes pairing, repeating the first parent鈥檚 phenotype.
  • \( Ad/aD \) or \( aD/Ad \) phenotypes replicate the \( F_1 \) generation's pattern.
  • \( aD/aD \) offspring showcase the phenotype of the original second parent.
By tracing these patterns, one can understand the inheritance dynamics of linked genes and predict future offspring genotypes and phenotypes.

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Most popular questions from this chapter

A rice breeder obtained a triple heterozygote carrying the three recessive alleles for albino flowers \((a l),\) brown awns \((b),\) and fuzzy leaves \((f u),\) all paired with their normal wild-type alleles. This triple heterozygote was testcrossed. The progeny phenotypes were \(170 \quad\) wild type \(150 \quad\) albino, brown, fuzzy \(5 \quad\) brown 3 albino, fuzzy \(710 \quad\) albino \(698 \quad\) brown, fuzzy \(42 \quad\) fuzzy \(38 \quad\) albino, brown a. Are any of the genes linked? If so, draw a map labeled with map distances. (Don't bother with a correction for multiple crossovers.) b. The triple heterozygote was originally made by crossing two pure lines. What were their genotypes?

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 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 \(10 \mathrm{m}\). \(u\). apart, what proportion of progeny will be \(A B / a b ?\)

The mother of a family with 10 children has blood type \(\mathrm{Rh}^{+}\). She also has a very rare condition (elliptocytosis, phenotype \(\mathrm{E}\) ) that causes red blood cells to be oval rather than round in shape but that produces no adverse clinical effects. The father is \(\mathrm{Rh}^{-}\) (lacks the \(\mathrm{Rh}^{+}\) antigen) and has normal red blood cells (phenotype e). The children are \(1 \mathrm{Rh}^{+} \mathrm{e}, 4 \mathrm{Rh}^{+} \mathrm{E},\) and \(5 \mathrm{Rh}^{-}\) e. Information is available on the mother's parents, who are \(\mathrm{Rh}^{+} \mathrm{E}\) and \(\mathrm{Rh}^{-}\) e. One of the 10 children (who is \(\mathrm{Rh}^{+} \mathrm{E}\) ) marries someone who is \(\mathrm{Rh}^{+} \mathrm{e}\), and they have an \(\mathrm{Rh}^{+} \mathrm{E}\) child. a. Draw the pedigree of this whole family. b. Is the pedigree in agreement with the hypothesis that the \(R h^{+}\) allele is dominant and \(R h^{-}\) is recessive? c. What is the mechanism of transmission of elliptocytosis? d. Could the genes governing the \(\mathrm{E}\) and \(\mathrm{Rh}\) phenotypes be on the same chromosome? If so, estimate the map distance between them, and comment on your result.

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