Problem 8 - Campbell Biology

Question

What is the probability that each of the following pairs of parents will produce the indicated offspring? (Assume independent assortment of all gene pairs.) $$\begin{array}{l}{\text { (a) } A A B B C C \times a a b b c c \rightarrow A a B b C c} \\ {\text { (b) } A A B b C c \times A a B b C c \rightarrow A A b b C C} \\ {\text { (c) } A a B b C c \times A a B b C c \rightarrow A a B b C c} \\ {\text { (d) } a a B b C C \times A A B b c c \rightarrow A a B b C c}\end{array}$$

Step-by-Step Solution

Answer

a) Probability = 1b) Probability = 1/32c) Probability = 1/8d) Probability = 1/2

Step 1 - Probability of Inheriting Each Gene

For each gene pair, determine the probability of the offspring inheriting the specific allele combination indicated. The genes are assumed to assort independently, so each gene pair's combination must be considered separately.

Step 2 - Calculate Probability for Part (a)

For part (a): - Parent genotypes: AA BB CC \times aa bb cc - Offspring genotype: Aa Bb CcEach gene pair results in a heterozygous offspring:- For gene A: AA \times aa = Aa (probability = 1)- For gene B: BB \times bb = Bb (probability = 1)- For gene C: CC \times cc = Cc (probability = 1).Multiplying the probabilities together: \[1 \times 1 \times 1 = 1\]

Step 3 - Calculate Probability for Part (b)

For part (b): - Parent genotypes: AA Bb Cc \times Aa Bb Cc - Offspring genotype: AA bb CCEach gene pair results in the probabilities:- For gene A: AA \times Aa = AA (probability = 1/2)- For gene B: Bb \times Bb = bb (probability = 1/4)- For gene C: Cc \times Cc = CC (probability = 1/4).Multiplying the probabilities together: \[ \frac{1}{2} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{32} \]

Step 4 - Calculate Probability for Part (c)

For part (c): - Parent genotypes: Aa Bb Cc \times Aa Bb Cc - Offspring genotype: Aa Bb CcEach gene pair results in the probabilities:- For gene A: Aa \times Aa = Aa (probability = 1/2)- For gene B: Bb \times Bb = Bb (probability = 1/2)- For gene C: Cc \times Cc = Cc (probability = 1/2).Multiplying the probabilities together: \[ \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \]

Step 5 - Calculate Probability for Part (d)

For part (d): - Parent genotypes: aa Bb CC \times AA Bb cc - Offspring genotype: Aa Bb CcEach gene pair results in the probabilities:- For gene A: aa \times AA = Aa (probability = 1)- For gene B: Bb \times Bb = Bb (probability = 1/2)- For gene C: CC \times cc = Cc (probability = 1).Multiplying the probabilities together: \[1 \times \frac{1}{2} \times 1 = \frac{1}{2} \]

Key Concepts

Independent Assortment

Independent assortment is a principle from genetics that explains how different genes are passed from parents to offspring. When genes assort independently, it means the inheritance of one gene does not influence the inheritance of another. This principle was discovered by Gregor Mendel through his work with pea plants. Independent assortment ensures that genetic variation occurs, as different combinations of alleles are possible for each offspring.

For example, in the given exercise, the genes A, B, and C assort independently. This allows us to calculate the probability of each gene combination separately and then multiply these probabilities to find the overall likelihood of the offspring’s genotype.

Allele Inheritance

Allele inheritance is the process through which alleles (different forms of a gene) are transmitted from parents to their offspring. Each parent provides one allele for each gene. The combination of these alleles determines the offspring's genotype.

In part (a) of the exercise, the parents have genotypes AA, BB, CC, and aa, bb, cc. The offspring's genotype is Aa, Bb, Cc. Each parent passes one allele per gene to the offspring:

• Gene A: One parent passes A, and the other passes a, resulting in Aa.
• Gene B: One parent passes B, and the other passes b, resulting in Bb.
• Gene C: One parent passes C, and the other passes c, resulting in Cc.

The overall probability is then found by multiplying the probabilities of each gene combination. This showcases how allele inheritance works.

Punnett Square

A Punnett Square is a graphical tool used to predict the genotypes of offspring from a particular cross. It helps visualize how alleles from parents combine during reproduction. To create a Punnett Square, list the alleles from one parent across the top and the alleles from the other parent down the side. Then fill in the squares by combining the alleles.

In part (b) of the exercise, for gene B, we use a Punnett Square to visualize the combination of Bb from both parents:

B | b

----

B | BB | Bb

b | Bb | bb

The probability of bb is 1 out of 4 squares, or 1/4. Similarly, independent Punnett Squares would be made for each gene pair to calculate the overall probability of the given genotype AA bb CC.

Genotype Probability

Genotype probability is the likelihood of an offspring having a particular genetic makeup. It can be calculated using the probabilities of inheritance for each gene. To find the overall probability, multiply the individual probabilities for each gene.

In part (c) of the exercise:

• For gene A, the probability of Aa is 1/2.
• For gene B, the probability of Bb is 1/2.
• For gene C, the probability of Cc is 1/2.

The overall probability of the genotype Aa Bb Cc is computed as follows:
\[\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \]
This method ensures accuracy when determining the likelihood of complex genotypic crosses in genetics.