The principles of segregation and independent assortment are important not only because they explain how heredity works, but also because they provide the means for predicting the outcome of genetic crosses. This predictive power has made genetics a powerful tool in agriculture and other fields, and the ability to apply the principles of heredity is an important skill for all students of genetics. Practice with genetics problems is essential for mastering the basic principles of heredity; no amount of reading and memorization can substitute for the experience gained by deriving solutions to specific problems in genetics.

You may find genetics problems difficult if you are unsure of where to begin or how to organize a solution to the problem. In genetics, every problem is different, and so no common series of steps can be applied to all genetics problems. Logic and common sense must be used to analyze a problem and arrive at a solution. Nevertheless, certain steps can facilitate the process, and solving the following problem will serve to illustrate those steps.

In mice, black coat color (B) is dominant over brown (b), and a solid pattern (S) is dominant over a white-spotted pattern (s). Color and spotting are controlled by genes that assort independently. A homozygous black, spotted mouse is crossed with a homozygous brown, solid mouse. All the F₁ mice are black and solid. A testcross is then carried out by mating the F₁ mice with brown, spotted mice.

Solution Strategy

What information is required in your answer to the problem?

First, determine what question or questions the problem is asking. Is it asking for genotypes, or genotypic ratios, or phenotypic ratios? This problem asks you to provide the genotypes of the parents and the F₁, the expected genotypes and phenotypes of the progeny of the testcross, and their expected proportions.

What information is provided to solve the problem?

Next, determine what information is provided that will be necessary for solving the problem. This problem gives important information about the dominance relations of the traits and the genes that encode them:

It is often helpful to write down the symbols at the beginning of the solution:

B—black S—solid

b—brown s—white-spotted

Next, write out the crosses, as given in the problem:

For help with this problem, review:

Sections 3.2 and 3.3.

Solution Steps

STEP 1: Write down any genetic information that can be determined from the phenotypes alone.

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From the phenotypes and the statement that they are homozygous, you know that the P-generation mice must be BB ss and bb SS. The F₁ mice are black and solid, both dominant traits, and so the F₁ mice must possess at least one black allele (B) and one solid allele (S). At this point, you may not be certain about the other alleles; so represent the genotype of the F₁ as B_ S_, where _ means that any allele is possible. The brown, spotted mice used in the testcross must be bb ss because both brown and spotted are recessive traits that will be expressed only if two recessive alleles are present. Record these genotypes on the crosses that you wrote out in step 2:

STEP 2: Break the problem down into smaller parts.

First, determine the genotype of the F₁. After this genotype has been determined, you can predict the results of the testcross and determine the genotypes and phenotypes of the progeny of the testcross. Second, because this cross includes two independently assorting loci, it can be conveniently broken down into two single-locus crosses: one for coat color and the other for spotting. Third, you can use a branch diagram to determine the proportion of progeny of the testcross with different combinations of the two traits.

STEP 3: Work the different parts of the problem.

Start by determining the genotype of the F₁ progeny. Mendel’s first law indicates that the two alleles at a locus separate, one going into each gamete. Thus, the gametes produced by the black, spotted parent contain B s and the gametes produced by the brown, solid parent contain b S, which combine to produce F₁ progeny with the genotype Bb Ss:

Use the F₁ genotype to work the testcross (Bb Ss × bb ss), breaking it into two single-locus crosses. First, consider the cross for coat color: Bb × bb. Any cross between a heterozygote and a homozygous recessive genotype produces a 1:1 phenotypic ratio of progeny (see Table 3.2):

Bb × bb

↓

½ Bb black

½ bb brown

Next, consider the cross for spotting: Ss × ss. This cross is between a heterozygote and a homozygous recessive genotype and produces ½ solid (Ss) and ½ spotted (ss) progeny (see Table 3.2):

Ss × ss

↓

½ Ss solid

½ ss spotted

Finally, determine the proportions of progeny with combinations of these characteristics by using the branch diagram.

STEP 4: Check all work.

As a last step, reread the problem, checking to see if your answers are consistent with the information provided. You have used the genotypes BB ss and bb SS in the P generation. Do these genotypes encode the phenotypes given in the problem? Are the F₁ progeny phenotypes consistent with the genotypes that you assigned? The answers are consistent with the information.