Question

when crossing two heterozygous individuals for brown eye color (bb x bb), the phenotypic ratio for the offspring includes a 75% chance of having ____ eyes. tongue rolling is considered a __ trait and shows a 50% probability in offspring when crossing a heterozygous roller (tt) with a non-roller (tt). if two heterozygous pea plants (rr) with yellow peas (r is dominant, r is recessive) are crossed, the predicted genotypic ratio is 1:2:1 for rr:rr:__. understanding punnett squares is crucial in genetic counseling to predict the likelihood of offspring inheriting ____ disorders.

Explanation:

First Question:

Step1: Recall Mendelian genetics for Bb x Bb cross.

The Punnett square for Bb (brown, dominant) x Bb:

	B	b
b	Bb	bb

Genotypes: BB (1/4), Bb (2/4), bb (1/4). Phenotypes: BB and Bb show brown (since B is dominant), bb shows blue (recessive). So brown phenotype: 3/4 = 75%.

Step2: Determine the eye color.

Since B is for brown (dominant), 75% have brown eyes.

Step1: Analyze Tt (roller) x tt (non - roller) cross.

Punnett square:

	T	t
t	Tt	tt

Genotypes: Tt (2/4), tt (2/4). Phenotype: Tt can roll tongue (dominant trait), tt can't. So 50% Tt (roller) and 50% tt (non - roller). So tongue rolling is a dominant trait.

Step2: Identify the trait type.

From the cross, the trait (tongue rolling) is dominant as T (roller) is dominant over t (non - roller).

Step1: Recall Rr x Rr cross for pea plants.

Punnett square for Rr (yellow, R dominant) x Rr:

	R	r
r	Rr	rr

Genotypes: RR (1/4), Rr (2/4), rr (1/4). So genotypic ratio RR:Rr:rr = 1:2:1.

Step2: Find the missing genotype.

From the ratio 1:2:1 (RR:Rr:__), the missing one is rr.

Answer:

brown

Second Question: