Question

modeling inheritance

• 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?

(there is a punnett square chart on the right with some handwritten genotypes: top row first cell ff, second cell ff; bottom row first cell ff, second cell ff (note: original handwritten might have variations but ocr focuses on text part))

Explanation:

Part A

Step1: Analyze Punnett Square

The Punnett square has parent genotypes (let's assume top parent is \( Ff \) and side parent is \( Ff \), but from the filled square and the rest, we see the offspring genotypes. The filled square and the others: looking at the Punnett square, the genotypes are \( Ff \) (three times? Wait, no, the square shown has one \( Ff \) (wait, the first square is \( Ff \)? Wait, the given Punnett square has: top row first square \( Ff \), second \( Ff \); bottom row first \( Ff \), second \( Ff \)? Wait, no, the user's image: the Punnett square has columns labeled \( F \) and \( F \)? Wait, no, maybe the parents are \( Ff \) and \( Ff \)? Wait, no, the top is \( F \) and \( f \)? Wait, maybe the parent on top is \( Ff \) (columns \( F \) and \( f \)) and parent on side is \( Ff \) (rows \( F \) and \( f \))? Wait, the filled square: first square (top left) is \( Ff \), then top right \( Ff \), bottom left \( Ff \), bottom right \( Ff \)? Wait, no, the user's handwritten: top row first \( Ff \), second \( Ff \); bottom row first \( Ff \), second \( Ff \). Wait, maybe the parents are \( Ff \) and \( Ff \), so the Punnett square gives offspring genotypes. Wait, the first square is filled as \( Ff \), then the rest: when you cross \( Ff \) (parent 1: \( F \) and \( f \)) and \( Ff \) (parent 2: \( F \) and \( f \)), the Punnett square is:

	\( F \)	\( f \)

| \( F \) | \( FF \)? Wait, no, the user's image: the columns are \( F \) and \( F \)? Wait, maybe the parent on top is \( FF \) and side is \( Ff \)? Wait, no, the first square is \( Ff \). Wait, maybe the top parent is \( F \) and \( f \) (so genotype \( Ff \)) and side parent is \( F \) and \( f \) (genotype \( Ff \)). Then the Punnett square:

• Top row (parent 1: \( F \), \( f \)) and side (parent 2: \( F \), \( f \)):

• Top left: \( F \times F = FF \)? No, the user's filled square is \( Ff \). Wait, maybe the top parent is \( F \) and \( f \) (genotype \( Ff \)) and side parent is \( F \) and \( f \) (genotype \( Ff \)), but the first square is \( Ff \), then the rest:

Wait, maybe the parents are \( Ff \) (mother) and \( Ff \) (father). Then the Punnett square:

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

But the user's filled square is \( Ff \) in first square. Wait, maybe the parents are \( Ff \) and \( Ff \), but the user's Punnett square has columns \( F \) and \( F \)? No, maybe the top parent is \( Ff \) (columns \( F \) and \( f \)) and side parent is \( Ff \) (rows \( F \) and \( f \)), but the user's handwritten shows all four squares as \( Ff \)? That can't be. Wait, maybe the parents are \( FF \) and \( Ff \). Let's re-express:

If parent 1 (top) is \( FF \) (genotype \( FF \), so gametes \( F \) and \( F \)) and parent 2 (side) is \( Ff \) (gametes \( F \) and \( f \)). Then the Punnett square:

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

But the first square (top left) is \( Ff \)? No, the user's first square is \( Ff \). Wait, maybe the parent on top is \( Ff \) (gametes \( F \) and \( f \)) and parent on side is \( FF \) (gametes \( F \) and \( F \)). Then:

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

Ah, that makes sense. So top parent: \( Ff \) (columns \( F \), \( f \)), side parent: \( FF \) (rows \( F \), \( F \)). Then the first square (top…

Step1: Determine Fur Color Dominance

Assume black fur is dominant (allele \( F \)) and white fur is recessive (allele \( f \)). The genotype \( Ff \) shows the dominant trait (black fur) because \( F \) is dominant over \( f \).

Step2: Calculate Percentage

There are 4 offspring, all with genotype \( Ff \). Since \( F \) is dominant, all 4 have black fur. The percentage is \( \frac{4}{4} \times 100 = 100\% \).

Step1: Determine Recessive Trait Expression

White fur is recessive, so it requires genotype \( ff \). From the offspring genotypes (\( Ff \) for all), none have \( ff \).

Step2: Calculate Percentage

Number of offspring with white fur: 0. Percentage is \( \frac{0}{4} \times 100 = 0\% \).

Answer:

\( Ff \) (all offspring have genotype \( Ff \))

Part B