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Chapter 3: Problem 14

Mendel crossed peas with round, green seeds with peas having wrinkled, yellow seeds. All \(\mathrm{F}_{1}\) plants had seeds that were round and yellow. Predict the results of testcrossing these \(F_{1}\) plants.

Short Answer

Expert verified
Based on Mendel's principles of segregation and independent assortment, a testcross between F1 generation pea plants (genotype RrYy, phenotype round yellow seeds) resulting from a cross between plants with round, green seeds (genotype RrYy) and plants with wrinkled, yellow seeds (genotype rrYY), will yield two potential phenotypes. These are: 1. Round and yellow seeds: This results from the dominant traits for round shape (R) and yellow color (Y). 2. Wrinkled and yellow seeds: This results from the recessive trait for wrinkled shape (r) and the dominant trait for yellow color (Y). Hence, no green seeds will be produced because the testcross parent is homozygous dominant for yellow (YY).

Step by step solution

01

Identify the genotypes of the parents

First, let's identify the genotypes of the parent plants. Round seeds are dominant (R) over wrinkled seeds (r), and yellow seeds are dominant (Y) over green seeds (y). The genotype of the pea plant with round, green seeds is RrYy since it passed on the gene for round seeds and one for green seeds. Likewise, the genotype of the pea plant with wrinkled, yellow seeds is rrYY since it passed on the gene for wrinkled seeds and one for yellow seeds.
02

Determine the F1 generation genotypes

Next, we'll determine the genotypes of the F1 generation by crossing the parent plants. We can use a Punnett square to visualize this cross: Parent 1: RrYy Parent 2: rrYY Punnett Square: ``` RY Ry rY ry +---------------- rY | RrYY RrYy rrYY rrYy rY | RrYY RrYy rrYY rrYy ``` So, the F1 generation plants all have the genotype RrYy, with round (R) and yellow (Y) seeds.
03

Perform the testcross

A testcross is done by crossing the F1 generation (RrYy) with a homozygous recessive genotype for both seed shape and seed color (rryy). Let's perform this testcross using another Punnett square: F1 plant: RrYy Testcross plant: rryy Punnett Square: ``` Ry rY Ry ry +---------------- ry | Rryy rryy Rryy rryy ry | Rryy rryy Rryy rryy ```
04

Predict the resulting phenotypes

Now, let's look at the resulting genotypes from the testcross and determine their corresponding phenotypes: 1. Rryy: Round and yellow seeds (dominant R and Y) 2. rryy: Wrinkled and yellow seeds (recessive r and dominant Y) Thus, the results of the testcross will produce offspring with either round and yellow seeds or wrinkled and yellow seeds. There will be no green seeds because the testcross parent is homozygous dominant for yellow (YY).

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Most popular questions from this chapter

In an intra-species cross performed in mustard plants of two different species (Brassicajuncea and Brassica oleracea), a tall plant \((T T)\) was crossed with a dwarf (tt) variety in each of the two species. The members of the \(\mathrm{F}_{1}\) generation were crossed to produce the \(\mathrm{F}_{2}\) generation. Of the \(\mathrm{F}_{2}\) plants, Brassica juncea had 60 tall and 20 dwarf plants, while Brassica oleracea had 100 tall and 20 dwarf plants. Use chi-square analysis to analyze these results.

Consider three independently assorting gene pairs, \(A / a, B / b,\) and \(C / c,\) where each demonstrates typical dominance \((A-, B-, C-)\) and recessiveness \((a a, b b, c c) .\) What is the probability of obtaining an offspring that is \(A A B b C c\) from parents that are \(A a B b C C\) and \(A A B b C c ?\)

Define critical \(p\) value. Explain what significance this value has for predicting the reproducibility of an experiment involving crosses. Explain why the null hypothesis is generally rejected for \(p\) values lower than 0.05

Early-onset myopia in humans is inherited as a simple dominant trait. Determine the genotypes of the parents and offspring for the following families. Mention alternate genotypes wherever applicable. (a) One normal (without early-onset myopia) parent and one abnormal (with early-onset myopia) parent produce six children, out of which only one is normal. (b) An abnormal male and a normal female produce five normal children.

Among dogs, short hair is dominant to long hair and dark coat color is dominant to white (albino) coat color. Assume that these two coat traits are caused by independently segregating gene pairs. For each of the crosses given below, write the most probable genotype (or genotypes if more than one answer is possible) for the parents. It is important that you select a realistic symbol set and define each symbol below. (a) dark, short \(\times\) dark, long \(26 \quad 24 \quad 0\) (b) albino, short \(\times\) albino, short \(0 \quad 0 \quad 102 \quad 33\) (c) dark, short \(\times\) albino, short \(16 \quad 0 \quad 16\) (d) dark, short \(\times\) dark, short \(175 \quad 67 \quad 61 \quad 21\) Assume that for cross (d), you were interested in determining whether fur color follows a 3: 1 ratio. Set up (but do not complete the calculations) a Chi-square test for these data [fur color in cross (d)].
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