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Chapter 5: Problem 31

A summer-squash plant that produces disc-shaped fruit is crossed with a summer-squash plant that produces long fruit. All the \(\mathrm{F}_{1}\) have disc-shaped fruit. When the \(\mathrm{F}_{1}\) are intercrossed, \(\mathrm{F}_{2}\) progeny are produced in the following ratio: \(9 / 16\) disc- shaped fruit : \(6 / 16\) spherical fruit \(: 1 / 16\) long fruit. Give the genotypes of the \(\mathrm{F}_{2}\) progeny.

Short Answer

Expert verified
Genotypes of the F2 generation: Disc - A-B-, Spherical - A-bb or aaB-, Long - aabb.

Step by step solution

01

Identify Dominance

The F1 generation, resulting from the cross of disc-shaped and long fruit plants, all have disc-shaped fruit. This indicates that disc-shaped is dominant over long fruit.
02

Consider Phenotypic Ratios

In the F2 generation, the ratio of 9 disc-shaped : 6 spherical : 1 long fruit suggests a modification of a 9:3:3:1 ratio in a dihybrid cross, which indicates involvement of two independently assorting genes.
03

Assign Alleles

Assume two gene pairs involved: - Gene A: Disc-shaped (A) is dominant over long (a), and - Gene B: Spherical (B) is dominant over long (b). Each shape is influenced by both gene pairs. Double recessive (aabb) results in long fruit, AaBb results in disc fruit, and aaBb (or Aabb) results in spherical fruit.
04

Genotype Analysis of F2 Generation

To understand the genotypes of individuals showing the phenotypes in the F2 generation: - **Disc-shaped**: The genotype combinations that result in this phenotype are A-B- (dominant alleles present). This can include AaBb, AABb, AaBB, etc. - **Spherical**: The genotype combinations are A-bb (like Aabb, AaBb) or aaB-. - **Long**: The genotype is aabb, the only combination resulting in long fruit.
05

Verify Genotype Distribution

Disc fruit results from 9/16 combinations (A-B-), spherical fruit from (3/16 A-bb + 3/16 aaB-) 6/16 combinations, and long fruit from 1/16 combination (aabb). Total combinations add up to 16, confirming our genotype assignment matches the phenotype ratio.

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Key Concepts

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

Phenotypic Ratios
Understanding phenotypic ratios is essential in genetic analysis, especially when dealing with dihybrid crosses. In this exercise, we observed the F2 generation resulting in a phenotypic ratio of 9:6:1. This particular ratio provides unique insights into the genetic interactions at play. Typically, a dihybrid cross results in a 9:3:3:1 phenotypic ratio, assuming simple dominance for two independent genes. However, the deviation here indicates an interaction between the genes that leads to different phenotypes.
In this scenario, the specific 9:6:1 ratio suggests an involvement of two gene pairs with dominant and recessive alleles interacting to affect the phenotype in a more complex manner.
  • The 9 represents progeny with at least one dominant allele from both gene pairs (disc-shaped).
  • The 6 comprises those with only one dominant allele or specific combinations leading to intermediate (spherical) forms.
  • The 1 refers to the recessive form appearing only when both pairs are fully recessive (long fruit).
This alteration in the expected phenotypic ratio highlights the importance of considering non-standard genetic interactions, which can result from multiple factors, including epistasis, where one gene masks or modifies the effect of another.
Genotype Analysis
Genotype analysis involves identifying the specific genetic makeup that leads to particular phenotypes. For the F2 progeny in the exercise, each phenotype corresponds to different genotypic combinations.
The overall objective of genotype analysis is to determine the alleles at each gene locus and how they contribute to observed traits. In the given problem:
  • Disc-shaped fruit: This phenotype results from dominant alleles in both gene pairs, denoted as A-B-. Possible genotypes include AaBb, AABb, or AaBB, among other combinations where at least one dominant allele from each gene is present.
  • Spherical fruit: This requires either one dominant allele from one gene and none from the other, leading to genotypes like A-bb or aaB-. Hence, genotypes such as Aabb or aaBb could result in spherical fruit.
  • Long fruit: As the fully recessive phenotype (1/16), this requires both pairs to be recessive, producing the aabb genotype.
Accurately analyzing these genotypic combinations not only reinforces comprehension of genetic inheritance but also accounts for any variation beyond simple Mendelian genetics.
Dominance in Genetics
Dominance in genetics determines how different alleles express themselves in an organism's phenotype. It's integral to understanding how certain traits show up in offspring, influencing both the F1 and F2 generations. In the problem at hand:
  • The fact that all F1 generation fruits were disc-shaped despite one parent having long fruit indicates that the disc-shaped allele is dominant over the allele for long fruit.
  • Dominance is not necessarily complete or simple. Variations, such as incomplete dominance or co-dominance, can also occur and lead to intermediate traits, but in this case, the dominance is more straightforward.
  • The dominance of the disc-shaped trait over long fruit is expressed when at least one dominant allele (A or B) is present.
Understanding dominance helps in predicting how traits will pass to subsequent generations and whether traits will be masked or expressed based on allele combinations. It remains a fundamental principle in comprehending genetic inheritance patterns beyond this exercise.

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Most popular questions from this chapter

Turkeys have black, bronze, or black-bronze plumage. Examine the results of the following crosses: $$ \begin{array}{ll} {\text { Parents }} & {\text { Offspring }} \\ \hline \text { Cross 1: black and bronze } & \text { all black } \\ \text { Cross 2: black and black } & 3 / 4 \text { black, } 1 / 4 \text { bronze } \\ \text { Cross 3: black-bronze and } & \text { all black-bronze } \\ \text { black-bronze } & \\ \text { Cross 4: black and bronze } & 1 / 2 \text { black, } 1 / 4 \text { bronze, } 1 / 4 \\ & \text { black-bronze } \\ \text { Cross 5: bronze and black- } & 1 / 2 \text { bronze, } 1 / 2 \text { black-bronze } \\ \text { bronze } & \\ \text { Cross 6: bronze and bronze } &\\\ &\text {4 bronze,} \text {1 / 4 black-bronze} \end{array} $$ Do you think these differences in plumage arise from incomplete dominance between two alleles at a single locus? If yes, support your conclusion by assigning symbols to each allele and providing genotypes for all turkeys in the crosses. If your answer is no, provide an alternative explanation and assign genotypes to all turkeys in the crosses.

In goats, a beard is produced by an autosomal allele that is dominant in males and recessive in females. We'll use the symbol \(B^{\mathrm{b}}\) for the beard allele and \(B^{+}\) for the beardless allele. Another independently assorting autosomal allele that produces a black coat \((W)\) is dominant to the allele for white coat \((w) .\) Give the phenotypes and their expected proportions for the following crosses. a. \(B^{+} B^{\mathrm{b}}\) Ww male \(\times B^{+} B^{\mathrm{b}}\) Ww female b. \(B^{+} B^{\mathrm{b}}\) Ww male \(\times B^{+} B^{\mathrm{b}}\) Ww female c. \(B^{+} B^{+}\) Ww male \(\times B^{\mathrm{b}} B^{\mathrm{b}}\) Ww female d. \(B^{+} B^{\mathrm{b}}\) Ww male \(\times B^{\mathrm{b}} B^{\mathrm{b}}\) Ww female

In unicorns, two autosomal loci interact to determine the type of tail. One locus controls whether a tail is present at all; the allele for a tail \((T)\) is dominant to the allele for tailless \((t)\). If a unicorn has a tail, then alleles at a second locus determine whether the tail is curly or straight. Farmer Baldridge has two unicorns with curly tails: when he crosses them, \(1 / 2\) of the progeny have curly tails, \(1 / 4\) have straight tails, and \(1 / 4\) do not have a tail. Give the genotypes of the parents and progeny in Farmer Baldridge's cross. Explain how he obtained the 2: 1: 1 phenotypic ratio in his cross.

In chickens, comb shape is determined by alleles at two loci \((R, r\) and \(P, p)\). A walnut comb is produced when at least one dominant allele \(R\) is present at one locus and at least one dominant allele \(P\) is present at a second locus (genotype \(\left.R_{-} P_{-}\right) .\) A rose comb is produced when at least one dominant allele is present at the first locus and two recessive alleles are present at the second locus (genotype \(\left.R_{-} p p\right)\). A pea comb is produced when two recessive alleles are present at the first locus and at least one dominant allele is present at the second (genotype \(r r P_{-}\) ). If two recessive alleles are present at the first and at the second loci \((r r p p)\), a single comb is produced. Progeny with what types of combs and in what proportions will result from the following crosses? a. \(R R \space P P \times \operatorname{rr}\space p p\) b. \(\operatorname{Rr} P p \times \operatorname{rrpp}\) c. \(\operatorname{Rr} P p \times \operatorname{Rr} P p\) d. \(\operatorname{Rr} p p \times \operatorname{Rr} p p\) e. \(\operatorname{Rr} p p \times \operatorname{rr} \mathrm{Pp}\) f. \(\operatorname{Rr} p p \times \operatorname{rrpp}\)

What is gene interaction? What is the difference between an epistatic gene and a hypostatic gene?
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