Question

1. the table below describes 100 offspring of the same two parents. what are the most likely genotypes of the parents? table showing fur - color and eye - color combinations with counts: black fur - black eyes: 22, black fur - white eyes: 27, red fur - black eyes: 25, red fur - white eyes: 26, total: 100 o ffee and ffee o ffee and ffee o ffee and ffee o ffee and ffee 4. a ffee mouse is bred to a ffee mouse. what is the probability that an offspring mouse will have black fur and red eyes? o 1/2 o 1/4 o 3/16 o 1/16

Explanation:

3.

Step1: Analyze offspring ratios

The offspring show a phenotypic ratio close to 9:3:3:1 which is characteristic of a di - hybrid cross (AaBb x AaBb). Here, FfEe x FfEe would result in such a ratio for two independently assorting genes.

Step2: Evaluate other options

Options A, B, and D would not produce a ratio close to what is seen in the offspring. For example, a cross like FFee and ffee would not give the variety of phenotypes in the observed ratios.

4.

Step1: Set up Punnett - square

A Ffee mouse has the genotype Ffee and a ffEe mouse has the genotype ffEe. For the fur - color gene (F/f), the cross is Ff x ff which gives Ff:ff = 1:1. For the eye - color gene (E/e), the cross is ee x Ee which gives Ee:ee = 1:1.

Step2: Calculate probability

To have black fur (F_) and red eyes (ee), we multiply the probabilities of each event. Probability of F_ is 1/2 and probability of ee is 1/2. So, \(P(F\_ee)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\).

Answer:

C. FfEe and FfEe