Question

figure 3
true or false:
a \ff\ parent and an \ff\ parent have the results shown in generation ii in figure 3.

Explanation:

Step1: Analyze Parent Genotypes

Parent I - 1 is \( ff \), Parent I - 2 has at least one \( F \) (since offspring have \( F\_ \)). Let's assume Parent I - 2 is \( Ff \) (to test the statement: one parent \( Ff \), one \( ff \)).

Step2: Predict Offspring Genotypes

Using Punnett square for \( Ff \times ff \):
| | \( f \) | \( f \) |
|---|---|---|
| \( F \) | \( Ff \) | \( Ff \) |
| \( f \) | \( ff \) | \( ff \) |

Offspring genotypes should be \( 50\% Ff \), \( 50\% ff \).

Step3: Compare with Generation II

Generation II has: II - 2 (\( ff \)), II - 1,3,4 (\( F\_ \), likely \( Ff \)). So 1 \( ff \), 3 \( Ff \)? Wait, no—wait, Parent I - 1 (\( ff \)) and I - 2 (\( Ff \)): number of offspring? The pedigree shows II - 2 (affected, \( ff \)), II - 1,3,4 (unaffected, \( F\_ \)). Wait, the cross \( Ff \times ff \) gives \( Ff \) (unaffected if \( f \) is recessive) and \( ff \) (affected). So in generation II, we have 1 \( ff \) (II - 2) and 3 \( Ff \) (II - 1,3,4). But the Punnett square for \( Ff \times ff \) has a 1:1 ratio. Wait, maybe the number of offspring isn't exactly 2, but in pedigrees, the ratio can be approximate. Wait, the statement is: "A 'Ff' parent and an 'ff' parent have the results shown in generation II". Let's check: Parent I - 1 is \( ff \), Parent I - 2: if the offspring include \( ff \) (II - 2) and \( Ff \) (II - 1,3,4), then yes—because \( Ff \times ff \) can produce \( Ff \) and \( ff \) offspring. The generation II has one \( ff \) (affected) and three \( Ff \) (unaffected), which is possible (since Punnett square is probability, not exact number). Wait, but let's re - evaluate:

Wait, the parent with \( ff \) (I - 1) and parent with \( Ff \) (I - 2). Their offspring: II - 2 is \( ff \) (so from \( f \) (I - 1) and \( f \) (I - 2)? No, wait, I - 1 is \( ff \) (so only \( f \) gametes), I - 2: if II - 2 is \( ff \), then I - 2 must have given an \( f \). So I - 2's genotype: to have II - 2 (\( ff \)), I - 2 must have an \( f \). And II - 1,3,4 are \( F\_ \), so I - 2 must have an \( F \). So I - 2 is \( Ff \). Then the cross is \( Ff \times ff \), which gives \( Ff \) (from \( F \) of I - 2 and \( f \) of I - 1) and \( ff \) (from \( f \) of I - 2 and \( f \) of I - 1). So the offspring can be \( Ff \) (unaffected) and \( ff \) (affected). In generation II, we have one \( ff \) (II - 2) and three \( Ff \) (II - 1,3,4). The Punnett square ratio is 1:1, but in a small sample (4 offspring), having 1 \( ff \) and 3 \( Ff \) is possible (due to chance). Wait, but the statement is whether a \( Ff \) and \( ff \) parent can have the results in generation II. Since the cross \( Ff \times ff \) can produce \( Ff \) and \( ff \) offspring, and generation II has both \( F\_ \) (likely \( Ff \)) and \( ff \) (II - 2), the answer is True? Wait, no—wait, maybe I messed up the dominance. Wait, the affected individuals are \( ff \) (recessive), so \( F \) is dominant, \( f \) is recessive. So \( Ff \) is unaffected, \( ff \) is affected. Parent I - 1: \( ff \) (affected, but in the pedigree, I - 1 is a square (male) and is black? Wait, no—wait, the first generation: I - 1 is a black square (male, \( ff \)), I - 2 is a white circle (female, \( F\_ \)). Then their offspring: II - 2 is a black circle (female, \( ff \)), II - 1,3,4 are white squares (males, \( F\_ \)). So the cross is \( ff \) (I - 1) and \( Ff \) (I - 2, because she has a child with \( ff \), so she must carry \( f \)). So the offspring genotypes: from \( ff \times Ff \), we get \( Ff \) (unaffected, white) and \( ff \) (affected, black). So in generation II, we have 1 \( ff \) (black) and 3 \( Ff \) (white). The Punnett square for \( Ff \times ff \) gives a 1:1 ratio of \( Ff \) to \( ff \). In a family with 4 children, having 3 \( Ff \) and 1 \( ff \) is within the realm of possible (since each child is an independent event, probabilities are for expected values, not exact counts). So the statement "A 'Ff' parent and an 'ff' parent have the results shown in generation II" is True, because the cross \( Ff \times ff \) can produce offspring with \( Ff \) (unaffected) and \( ff \) (affected), which matches the genotypes in generation II (II - 1,3,4: \( F\_ \) ( \( Ff \) ), II - 2: \( ff \) ).

Answer:

True