Question

What is the chance that a dihybrid cross will produce a homozygous recessive in both traits? a. \(9 / 16\) b. \(1 / 4\) c. \(1 / 16\) d. \(3 / 16\) e. \(1 / 8\)

Short answer

The chance that a dihybrid cross will produce a homozygous recessive in both traits is c. \(1 / 16\).

Step-by-step solution

Determine the parental genotypes for each trait

In a dihybrid cross, the parents are both heterozygous for both traits. So, for our example, let Trait 1 be represented by "A" (dominant) and "a" (recessive) and Trait 2 be represented by "B" (dominant) and "b" (recessive). The parental genotypes will be AaBb x AaBb.

Create Punnett squares for each trait

We will create a Punnett square for each trait to determine the probability of getting a homozygous recessive genotype for each. Trait 1: | | A | a | |---|---|---| | A | AA| Aa| | a | Aa| aa| Trait 2: | | B | b | |---|---|---| | B | BB| Bb| | b | Bb| bb| From the Punnett squares, we find the probability of getting a homozygous recessive genotype for Trait 1 (aa) is 1/4 and for Trait 2 (bb) is also 1/4.

Calculate the combined probability

To find the combined probability of getting a homozygous recessive genotype for both traits (aabb), we multiply the probabilities of the individual traits: Probability (Trait 1) x Probability (Trait 2) = \(\frac{1}{4}\) x \(\frac{1}{4}\)= \(\frac{1}{16}\) Therefore, the chance that a dihybrid cross will produce a homozygous recessive in both traits is 1/16. The correct answer is c. \(1 / 16\).

Key concepts

Homozygous Recessive

When discussing genetics, the term "homozygous recessive" refers to an individual who carries two copies of the recessive allele for a specific trait. Alleles are different forms of a gene, and they can be dominant or recessive.
Dominant alleles are represented by uppercase letters, whereas recessive alleles are indicated by lowercase letters. For a trait to be expressed as recessive, both alleles must be recessive, hence the term homozygous recessive.

• For example, if "a" is the recessive allele, a homozygous recessive genotype would be "aa".

Homozygous recessive genotypes are essential in understanding many genetic patterns and predictions. Only individuals with a homozygous recessive genotype will show the recessive trait when it is masked by a dominant allele in heterozygotes (Aa or Bb and so forth). This concept is especially important when predicting outcomes in genetic cross experiments, like a dihybrid cross.

Punnett Square

The Punnett square is a visual tool used in genetics to predict the potential genotypes of offspring from a cross between two parents. Invented by Reginald Punnett, it's a simple grid that makes understanding genetic probability easy and intuitive.

• To create a Punnett square, list the alleles of one parent on the top and the alleles of the other parent on the side, forming a grid where their intersections display possible allele combinations.
• In a dihybrid cross, you consider two traits simultaneously, usually denoted as "AaBb x AaBb" for parents, leading to a 4x4 grid.

Each cell in the grid represents a potential genotype of the offspring from this genetic combination. Observing the squares helps determine the probability of each genotype occurring, including the homozygous recessive, homozygous dominant, and heterozygous combinations. Thus, Punnett squares are fundamental in solving genetics problems as they visually simplify complex genetic inheritance patterns.

Genotype Probability

Genotype probability refers to the likelihood or chance of a particular genetic composition occurring in the offspring of a genetic cross. This understanding is crucial when analyzing genetic outcomes, as it helps in predicting traits not yet observed.

• Probabilities are often calculated as fractions or percentages.
• In our context, using the Punnett square, you'd see that each box implies a specific genotype probability assuming each parental allele combination has an equal chance of occurring.

For example, in a dihybrid cross of "AaBb x AaBb", when calculating for a homozygous recessive genotype "aabb", you first determine the probability for each trait separately: the chance of getting "aa" is \( \frac{1}{4} \), and for "bb" it is also \( \frac{1}{4} \). You then multiply these probabilities: \[ \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} \] which gives the overall probability of that genotype in the offspring. This probability helps geneticists and students alike determine the expected distribution of traits in future generations.