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Chapter 4: Problem 4

The trait of medium-sized leaves in iris is determined by the genetic condition \(P P^{\prime}\). Plants with large leaves are \(P P\), whereas plants with small leaves are \(P^{\prime} P^{\prime} .\) A cross is made between two plants each with medium-sized leaves. If they produce 80 seedlings, what would be the expected phenotypes, and in what numbers would they be expected? What is the term for this allelic relationship?

Short Answer

Expert verified
Answer: The expected phenotypes are 20 large leaves, 40 medium-sized leaves, and 20 small leaves. This allelic relationship is known as "incomplete dominance" or "partial dominance."

Step by step solution

01

Set up the Punnett square

First, we need to set up the Punnett square for both parent plants with medium-sized leaves. Their genetic makeup would be \(P P^{\prime}\).
02

Fill in the Punnett square

Each cell of the Punnett square should be filled with the results of combining the letters from each parent. Write the dominant allele "P" first in each cell, followed by the recessive allele "P'". $$ \begin{array}{cc|cc} & & \textbf{Parent 1} & \\ & & P & P^{\prime} \\ \cline{2-4} \textbf{Parent 2} & P & PP & PP^{\prime} \\ & P^{\prime} & PP^{\prime} & P^{\prime} P^{\prime} \end{array} $$
03

Calculate expected phenotypic ratios

Now we need to analyze the results of the Punnett square to find the phenotypic ratios. Use the given genotype information to determine the phenotype of each genotype combination. 1. \(PP\): Large leaves 2. \(PP^{\prime}\): Medium-sized leaves 3. \(P^{\prime}P^{\prime}\): Small leaves We can see in the Punnett square that there is 1 \(PP\), 2 \(PP^{\prime}\), and 1 \(P^{\prime}P^{\prime}\). The ratio is 1:2:1 for large : medium : small leaves.
04

Determine the expected numbers of each phenotype

There are 80 seedlings produced, so apply the phenotypic ratio to the total of 80 seedlings: 1. Large leaves: \(80 \times \frac{1}{4} = 20\) 2. Medium-sized leaves: \(80 \times \frac{2}{4} = 40\) 3. Small leaves: \(80 \times \frac{1}{4} = 20\) The expected numbers of each phenotype are 20 large leaves, 40 medium-sized leaves, and 20 small leaves.
05

Identify the allelic relationship term

The term for this allelic relationship is "incomplete dominance" or "partial dominance," as neither allele is completely dominant over the other, resulting in an intermediate phenotype (medium-sized leaves) when both alleles are present in the offspring.

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Most popular questions from this chapter

The specification of the anterior-posterior axis in Drosophila embryos is initially controlled by various gene products that are synthesized and stored in the mature egg following oogenesis. Mutations in these genes result in abnormalities of the axis during embryogenesis, illustrating maternal effect. How do such mutations vary from those involved in organelle heredity that illustrate extranuclear inheritance? Devise a set of parallel crosses and expected outcomes involving mutant genes that contrast maternal effect and organelle heredity.

In four o'clock plants, many flower colors are observed. In a cross involving two true-breeding strains, one crimson and the other white, all of the \(\mathrm{F}_{1}\) generation were rose color. In the \(\mathrm{F}_{2}\) four new phenotypes appeared along with the \(P_{1}\) and \(F_{1}\) parental colors. The following ratio was obtained: \(1 / 16\) crimson \(4 / 16\) rose \(2 / 16\) orange \(\quad 2 / 16\) pale yellow 1/16 yellow \(\quad 4 / 16\) white \(2 / 16\) magenta Propose an explanation for the inheritance of these flower colors.

The maternal-effect mutation bicoid (bcd) is recessive. In the absence of the bicoid protein product, embryogenesis is not completed. Consider a cross between a female heterozygous for the bicoid mutation \(\left(b c d^{+} / b c d^{-}\right)\) and a homozygous male \(\left(b c d^{\left.-/ b c d^{-}\right)}\right.\) (a) How is it possible for a male homozygous for the mutation to exist? (b) Predict the outcome (normal vs, failed embryogenesis) in the \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) generations of the cross described.

In goats, development of the beard is due to a recessive gene. The following cross involving true-breeding eoats was made and carried to the \(\mathrm{F}_{2}\) generation: \(\mathrm{P}_{1}=\) bearded female \(\times\) beardless male \(\mathrm{F}_{1}:\) all bearded males and beardless females \\[ \mathbf{F}_{1} \times \mathbf{F}_{1} \rightarrow\left\\{\begin{array}{l} 1 / 8 \text { beardless males } \\ 3 / 8 \text { bearded males } \\ 3 / 8 \text { beardless females } \\ 1 / 8 \text { bearded females } \end{array}\right. \\] Offer an explanation for the inheritance and expression of this trait, diagramming the cross. Propose one or more crosses to test your hypothesis.

In cattle, coats may be solid white, solid black, or black-andwhite spotted. When true-breeding solid whites are mated with true-breeding solid blacks, the \(\mathrm{F}_{1}\), generation consists of all solid white individuals. After many \(\mathrm{F}_{1} \times \mathrm{F}_{1}\) matings, the following ratio was observed in the \(\mathrm{F}_{2}\) generation: \(12 / 16\) solid white \(3 / 16\) black-and-white spotted \(1 / 16\) solid black Rxplain the mode of inheritance governing coat color by determining how many gene pairs are involved and which genotypes yield which phenotypes. Is it possible to isolate a true-breeding strain of black-and-white spotted cattle? If so, what genotype would they have? If not, explain why not.
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