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Chapter 3: Problem 13

How many different types of gametes can be formed by individuals of the following genotypes? What are they in each case? (a) \(A a B b\) (b) \(A a B B\) (c) \(A a B b C c\) (d) \(A a B B c c\) (e) \(A a B b c c,\) and (f) \(A a B b C c D d E e ?\)

Short Answer

Expert verified
(a) Genotype: AaBb Answer: 4 gametes: AB, Ab, aB, and ab. (b) Genotype: AaBB Answer: 2 gametes: AB and aB. (c) Genotype: AaBbCc Answer: 8 gametes: ABC, ABc, AbC, Abc, aBC, aBc, abC, and abc. (d) Genotype: AaBBcc Answer: 2 gametes: ABc and aBc. (e) Genotype: AaBbcc Answer: 4 gametes: ABc, Abc, aBc, and abc. (f) Genotype: AaBbCcDdEe Answer: 32 gametes. The different combinations of alleles can be determined systematically using the law of independent assortment.

Step by step solution

01

(a) Genotype: AaBb

There are two possibilities for each gene: A or a and B or b. To determine the number of gametes, we can use the formula \(2^n\), where n is the number of heterozygous gene pairs. Here, n=2. Number of gametes = \(2^2 = 4\) The possible gametes are: AB, Ab, aB, and ab.
02

(b) Genotype: AaBB

There are two possibilities for the first gene: A or a and only one possibility for the second gene: B. In this case, n=1. Number of gametes = \(2^1 = 2\) The possible gametes are: AB and aB.
03

(c) Genotype: AaBbCc

There are two possibilities for each gene: A or a, B or b, and C or c. Here, n=3. Number of gametes = \(2^3 = 8\) The possible gametes are: ABC, ABc, AbC, Abc, aBC, aBc, abC, and abc.
04

(d) Genotype: AaBBcc

There are two possibilities for the first gene: A or a, one possibility for the second gene: B, and one possibility for the third gene: c. In this case, n=1. Number of gametes = \(2^1 = 2\) The possible gametes are: ABc and aBc.
05

(e) Genotype: AaBbcc

There are two possibilities for the first gene: A or a and two possibilities for the second gene: B or b. Here, n=2. Number of gametes = \(2^2 = 4\) The possible gametes are: ABc, Abc, aBc, and abc.
06

(f) Genotype: AaBbCcDdEe

There are two possibilities for each of the five genes, meaning n=5. Number of gametes = \(2^5 = 32\) The possible gametes can be systematically determined using the law of independent assortment, resulting in 32 different combinations of alleles.

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Most popular questions from this chapter

Why was the garden pea a good choice as an experimental organism in Mendel's work?

Distinguish between homozygosity and heterozygosity.

The basis for rejecting any null hypothesis is arbitrary. The researcher can set more or less stringent standards by deciding to raise or lower the critical \(p\) value. Would the use of a standard of \(p=0.10\) be more or less stringent in failing to reject the null hypothesis? Explain.

To assess Mendel's law of segregation using tomatoes, a true- breeding tall variety (SS) is crossed with a true-breeding short variety \((s s) .\) The heterozygous tall plants \((S s)\) were crossed to produce the two sets of \(\mathrm{F}_{2}\) data as follows: $$\begin{array}{cc} \text { Set I } & \text { Set II } \\ 30 \text { tall } & 300 \text { tall } \\ 5 \text { short } & 50 \text { short } \end{array}$$ (a) Using chi-square analysis, analyze the results for both datasets. Calculate \(\chi^{2}\) values, and estimate the \(p\) values in both cases. (b) From the analysis in part (a), what can you conclude about the importance of generating large datasets in experimental settings?

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.
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