Question

orange fish and yellow fish make striped babies. a yellow fish and a striped fish mate. what are the phenotypes of the babies?
○ 100% striped fish
○ 50% striped, 50% yellow
○ 100% yellow fish
○ 50% striped, 25% yellow, 25% orange

question 8
1 pts
a long - tailed tiger and a short - tailed tiger make tiger babies with medium length tails. what parents would be needed to make 25% long - tailed tigers, 50% medium - tailed tigers, and 25% short - tailed tigers?
○ ls x ls
○ ll x ss
○ ll x ls
○ ss x ls

question 9
1 pts
a woman who is heterozygous for blood type b has a child with someone who is blood type ab. what are the phenotypes of their children?
○ 25% ab, 50% b, 25% o
○ 50% ab, 25% b, 25% o
○ 50% ab, 50% b
○ 25% ab, 25% a, 50% b

Explanation:

Question about Fish Phenotypes:

Orange and yellow fish make striped babies, so striped is a blend (maybe codominance or incomplete dominance). A yellow (let's say genotype \( Y \)) and striped (genotype \( OS \), where \( O \) is orange, \( S \) is striped? Wait, no—first cross: orange + yellow = striped, so striped is a combination. Then yellow (\( Y \)) and striped (\( OS \)): gametes from yellow: \( Y \); from striped: \( O \), \( S \). Wait, no, maybe the first cross: orange (\( O \)) and yellow (\( Y \)) produce striped (\( OY \)). Then yellow (\( Y \)) and striped (\( OY \)): offspring would be \( Y \times O \) (YO, striped) and \( Y \times Y \) (YY, yellow)? No, the options: 100% Striped? No. Wait, the first cross: orange + yellow = striped (so striped is heterozygous, maybe \( OY \)). Then yellow (\( YY \)) and striped (\( OY \)): cross \( YY \times OY \). Gametes: \( Y \) (from YY) and \( O, Y \) (from OY). Offspring: \( YO \) (striped) and \( YY \) (yellow), each 50%? Wait the options: 50% Striped, 50% Yellow. Wait the first part: "Orange fish and yellow fish make striped babies"—so orange (\( O \)) and yellow (\( Y \)) → striped (\( OY \)). Then "A yellow fish and a striped fish mate"—yellow is \( YY \), striped is \( OY \). Cross: \( YY \times OY \). Punnett square:

	O	Y
Y	OY (striped)	YY (yellow)

So 50% striped, 50% yellow. So the answer is "50% Striped, 50% Yellow".

Long-tailed (LL), medium (LS), short (SS). To get 25% LL, 50% LS, 25% SS, the cross should be LS × LS (since \( LS \times LS \) gives \( LL \) (25%), \( LS \) (50%), \( SS \) (25%)). Wait the question is "What parents would be needed to make 25% long - tailed tigers, 50% medium - tailed tigers, and 25% short - tailed tigers?" The options: LS × LS. Let's check: \( LS \) (medium) × \( LS \) (medium). Gametes: L and S from each. Punnett square:

	L	S
S	LS (medium, 50%)	SS (short, 25%)	Wait no, that's 25% LL, 50% LS, 25% SS? Wait no, LL is 25% (1/4), LS is 2/4 (50%), SS is 1/4 (25%). Wait the question says "25% long - tailed, 50% medium - tailed, 25% short - tailed". So the cross is LS × LS. So the answer is LS × LS.

Woman: heterozygous for B, so genotype \( I^B i \) (since blood type B heterozygous: \( I^B i \), because \( I^B I^B \) is homozygous B). Partner: AB, genotype \( I^A I^B \). Cross: \( I^B i \times I^A I^B \). Gametes from woman: \( I^B \), \( i \); from partner: \( I^A \), \( I^B \). Punnett square:

	\( I^A \)	\( I^B \)
\( i \)	\( I^A i \) (A, 25%)	\( I^B i \) (B, 25%)	Wait no, wait: Wait the woman is heterozygous for B: so \( I^B i \) (blood type B, since \( I^B \) is dominant over \( i \)). Partner is AB: \( I^A I^B \). So gametes: woman: \( I^B \), \( i \); partner: \( I^A \), \( I^B \). Offspring:

• \( I^A I^B \): AB (25%)
• \( I^B I^B \): B (25%)
• \( I^A i \): A (25%)
• \( I^B i \): B (25%)

Wait the options: "25% AB, 50% B, 25% A"—wait no, my calculation: AB (25%), B (50%: \( I^B I^B \) and \( I^B i \)), A (25%: \( I^A i \)). Wait the options: "25% AB, 50% B, 25% A"—yes, that's the case. Wait the options given: "25% AB, 50% B, 25% A" (wait the last option: "25% AB, 25% A, 50% B"). Let's recalculate:
Woman: \( I^B i \) (gametes: \( I^B \), \( i \))
Partner: \( I^A I^B \) (gametes: \( I^A \), \( I^B \))
Offspring:

1. \( I^A I^B \): AB (25%)
2. \( I^B I^B \): B (25%)
3. \( I^A i \): A (25%)
4. \( I^B i \): B (25%)

So AB: 25%, B: 50% (25 + 25), A: 25%. So the option is "25% AB, 50% B, 25% A" (the last option: "25% AB, 25% A, 50% B").

Answer:

50% Striped, 50% Yellow

Question 8 (Tiger Tails):