Question

parent 1 is heterozygous for dimples while parent 2 is homozygous recessive for no dimples.
d = dimples, d = no dimples
based on the information and the punnett square, select the true statement(s) about the monohybrid genetic cross.
0% of offspring will have dimples.
the genotype for offspring with no dimples is dd.
50% of offspring will have dimples.
there is a 4 in 4 chance the offspring will have dimples.
the offspring with dimples are heterozygous for the trait.
100% of offspring will have no dimples.

Explanation:

To solve this, we analyze the genetic cross:

• Parent 1 (heterozygous for dimples): Genotype \( Dd \) (produces gametes \( D \) and \( d \)).
• Parent 2 (homozygous recessive for no dimples): Genotype \( dd \) (produces gametes \( d \) and \( d \)).

Step 1: Create the Punnett Square

	\( D \)	\( d \)
\( d \)	\( Dd \)	\( dd \)

Step 2: Analyze Each Statement

1. 0% of offspring will have dimples:

Offspring with \( Dd \) (dimples) are 2 out of 4. So this is false.

1. The genotype for offspring with no dimples is \( dd \):

No dimples is recessive, so only \( dd \) shows this trait. True.

1. 50% of offspring will have dimples:

\( Dd \) (dimples) is 2/4 = 50%. True.

1. There is a 4 in 4 chance the offspring will have dimples:

Only 2/4 have dimples. False.

1. The offspring with dimples are heterozygous for the trait:

Dimpled offspring have \( Dd \) (one dominant, one recessive allele) → heterozygous. True.

1. 100% of offspring will have no dimples:

50% have dimples (\( Dd \)). False.

1. There is a 2 in 4 chance the offspring will have no dimples:

\( dd \) (no dimples) is 2/4. True.

Answer:

The true statements are:

• The genotype for offspring with no dimples is \( dd \).
• 50% of offspring will have dimples.
• The offspring with dimples are heterozygous for the trait.
• There is a 2 in 4 chance the offspring will have no dimples.

(In boxed form for each correct option’s identifier, but since options are checkboxes, list the true ones as above.)