Question

activity c: modeling inheritance get the gizmo ready: - click clear. - drag a black mouse and a white mouse into the parent boxes. question: how do scientists predict the genotypes of offspring? 1. model: scientists use a punnett square to model the different possible offspring genotypes from a parent pair. the parent genotypes are written across the top and side of the square, as shown. the four possible offspring genotypes are then filled in. the first square is filled in for you. fill in the remaining squares. a. what are the genotypes of the offspring? ______ b. what percentage of the offspring will have black fur? ____ c. what percentage of the offspring will have white fur? ______

Answer:

Part 1: Filling the Punnett Square

To fill the Punnett square, we use the parent genotypes. From the given square, the top row has "F" and "F" (assuming the first parent is \( FF \) or one parent has \( F \) alleles across the top, and the side has "f" and "f" (the other parent has \( f \) alleles down the side). Wait, actually, looking at the first filled square (top - left) as \( Ff \), let's analyze:

The top of the Punnett square (horizontal) represents the alleles from one parent, and the side (vertical) represents the alleles from the other parent. Let's assume the top parent has alleles \( F \) and \( F \) (so across the top: \( F \) (first column) and \( F \) (second column)), and the side parent has alleles \( f \) and \( f \) (down the side: \( f \) (first row) and \( f \) (second row)).

• Top - left square: \( F \) (from top) and \( f \) (from side) → \( Ff \) (already filled)
• Top - right square: \( F \) (from top) and \( f \) (from side) → \( Ff \)
• Bottom - left square: \( F \) (from top) and \( f \) (from side) → \( Ff \)
• Bottom - right square: \( F \) (from top) and \( f \) (from side) → \( Ff \)

Wait, maybe the top parent is \( Ff \)? No, the first square is \( Ff \). Wait, perhaps the top parent has alleles \( F \) (first column) and \( F \) (second column), and the side parent has \( f \) (first row) and \( f \) (second row). So:

• Top - left: \( F \times f = Ff \) (filled)
• Top - right: \( F \times f = Ff \)
• Bottom - left: \( F \times f = Ff \)
• Bottom - right: \( F \times f = Ff \)

Wait, maybe the parents are \( Ff \) and \( ff \)? No, let's re - examine. Alternatively, if the top parent is \( F \) (allele 1) and \( F \) (allele 2), and the side parent is \( f \) (allele 1) and \( f \) (allele 2), then all four squares will be \( Ff \). But let's confirm with the standard Punnett square rules.

Part A: Genotypes of the offspring

From the filled Punnett square (after filling all squares), the genotypes of the offspring are all \( Ff \)? Wait, no, maybe I made a mistake. Wait, if one parent is \( FF \) (homozygous dominant) and the other is \( ff \) (homozygous recessive), then the Punnett square would be:

	\( F \)	\( F \)
\( f \)	\( Ff \)	\( Ff \)

So all four squares are \( Ff \). Wait, but maybe the parents are \( Ff \) and \( Ff \)? No, the first square is \( Ff \). Wait, the problem says "drag a black mouse and a white mouse into the parent boxes". Let's assume black is dominant (\( F \)) and white is recessive (\( f \)). So a black mouse could be \( FF \) or \( Ff \), and a white mouse is \( ff \). If the white mouse is \( ff \) (homozygous recessive) and the black mouse is \( FF \) (homozygous dominant), then the cross is \( FF \times ff \).

Step 1: Determine the alleles of parents

• Black mouse (let's say dominant, \( F \)): If it's homozygous dominant, alleles are \( F \) and \( F \).
• White mouse (recessive, \( f \)): alleles are \( f \) and \( f \).

Step 2: Construct the Punnett square

	\( F \)	\( F \)
\( f \)	\( Ff \)	\( Ff \)

So the four offspring genotypes are all \( Ff \). Wait, but maybe the black mouse is \( Ff \)? Let's check. If black mouse is \( Ff \) (heterozygous) and white mouse is \( ff \) (homozygous recessive), the cross is \( Ff \times ff \):

	\( F \)	\( f \)
\( f \)	\( Ff \)	\( ff \)

But the first filled square is \( Ff \), so if the top parent is \( Ff \) (with alleles \( F \) and \( f \) across the top) and the side parent is \( ff \) (with alleles \( f \) and \( f \) down the side), then:

• Top - left: \( F \times f = Ff \) (filled)
• Top - right: \( f \times f = ff \)
• Bottom - left: \( F \times f = Ff \)
• Bottom - right: \( f \times f = ff \)

But the problem's first filled square is \( Ff \), and the side has two "f"s? Wait, the original Punnett square in the image has top (horizontal) with two "F"s? Wait, the user's image shows the Punnett square with top labels "F" and "F" (two Fs) and side labels "f" and "f" (two fs). So:

• Top - left: \( F \) (top) and \( f \) (side) → \( Ff \) (filled)
• Top - right: \( F \) (top) and \( f \) (side) → \( Ff \)
• Bottom - left: \( F \) (top) and \( f \) (side) → \( Ff \)
• Bottom - right: \( F \) (top) and \( f \) (side) → \( Ff \)

So the remaining squares:

• Top - right: \( Ff \)
• Bottom - left: \( Ff \)
• Bottom - right: \( Ff \)

Part A: Genotypes of the offspring

From the filled Punnett square (after filling all four squares), the genotypes of the offspring are all \( Ff \)? Wait, no, if the top parent is \( FF \) and the side parent is \( ff \), then all offspring are \( Ff \). So the genotypes of the offspring are \( Ff \) (all four, so the genotypes are \( Ff \) (since all squares are \( Ff \))? Wait, no, if the Punnett square has four squares, all with \( Ff \), then the offspring genotypes are \( Ff \) (100% of them have genotype \( Ff \)).

Part B: Percentage of offspring with black fur

Black fur is dominant (assuming \( F \) is the dominant allele for black fur, and \( f \) is recessive for white fur). The genotype \( Ff \) has the dominant allele \( F \), so it will express black fur. Since all offspring have genotype \( Ff \), the percentage of offspring with black fur is \( \frac{4}{4} \times 100\% = 100\% \).

Part C: Percentage of offspring with white fur

White fur is recessive, so it is expressed only when the genotype is \( ff \). Since none of the offspring have genotype \( ff \) (all are \( Ff \)), the percentage of offspring with white fur is \( \frac{0}{4} \times 100\% = 0\% \).

A. Genotypes of the offspring

All four offspring have the genotype \( Ff \). So the genotypes of the offspring are \( Ff \) (or we can say the possible genotypes are \( Ff \), and all offspring will have \( Ff \) as their genotype).

B. Percentage of offspring with black fur

Since all offspring have the dominant allele (in genotype \( Ff \)) for black fur, the percentage is \( 100\% \) (or \( \frac{4}{4}\times100 = 100\% \)).

C. Percentage of offspring with white fur

Since none of the offspring have the recessive - only genotype (\( ff \)) for white fur, the percentage is \( 0\% \) (or \( \frac{0}{4}\times100 = 0\% \)).

Final Answers:

A. The genotypes of the offspring are \( Ff \) (all offspring have genotype \( Ff \)).
B. The percentage of offspring with black fur is \( 100\% \).
C. The percentage of offspring with white fur is \( 0\% \).

(Note: If the parents were different (e.g., black mouse is \( Ff \) and white mouse is \( ff \)), the answers would change. But based on the first filled square being \( Ff \) and the Punnett square structure, the most likely cross is \( FF \times ff \), leading to these answers. If the black mouse was \( Ff \), the cross \( Ff \times ff \) would give \( Ff \) and \( ff \) genotypes, but the first filled square is \( Ff \), and the side has two "f"s, so the top parent likely has two "F"s (homozygous dominant) and the side parent has two "f"s (homozygous recessive).)