Question

← zebra mussel genetics
(some text about zebra mussels, their introduction to the united states, reproduction, and shell color phenotypes: dark (genotype aa), zebra - striped (genotype aa), and solid light - colored (genotype aa) which are not hardy in cold water temperatures. there is an image of three mussels with different shell patterns. then a question: if you cross a mussel with dark, zebra - striped shells with a mussel with solid light - colored shells, how many offspring would you expect to have the homozygous dark - colored shells? with multiple - choice options a: 0%, b: 50%, c: 75%, d: 100%. another question: which cross would produce the greatest genetic diversity in the offspring? (text partially visible))

Explanation:

Step1: Determine Genotypes

Let's assume the dark, zebra - striped mussel has genotype \(AW\) (since it's zebra - striped, which is a heterozygous trait from the context) and the mussel with solid light - colored shells has genotype \(aa\) (wait, actually, from the problem, the solid light - colored shells are genotype \(aa\)? Wait, no, re - reading: the solid light - colored shells (genotype \(aa\))? Wait, the dark, zebra - striped is \(AW\), and the solid light - colored is \(aa\)? Wait, no, the problem says: "heterozygous individuals (Aa) have solid, dark - colored shells" – wait, maybe I misread. Wait, the zebra - striped is \(AW\), and the solid light - colored is \(aa\)? Wait, no, let's re - analyze.

Wait, the problem: "If you cross a mussel with dark, zebra - striped shells (genotype \(AW\)) with a mussel with solid light - colored shells (genotype \(aa\)), how many offspring would you expect to have the same dark, zebra - striped shells?"

Wait, no, maybe the genotypes are: Let's assume that the zebra - striped (dark, zebra - striped) is a heterozygous genotype, say \(Ww\) (but the problem uses \(AW\) or \(Aa\)? Wait, the text says "heterozygous individuals (Aa) have solid, dark - colored shells" – oh, maybe I mixed up. Wait, the zebra - striped is a different genotype. Wait, the problem is about zebra mussel genetics. Let's assume that the dark, zebra - striped mussel has genotype \(Ww\) (where \(W\) is for zebra - striped and \(w\) is for non - striped, but maybe the solid light - colored is \(ww\)? No, this is getting confusing. Wait, maybe the cross is between \(Aa\) (zebra - striped) and \(aa\) (solid light - colored). Wait, no, the question is: cross a mussel with dark, zebra - striped shells (let's say genotype \(Aa\)) with a mussel with solid light - colored shells (genotype \(aa\)). Wait, no, the key is to do a Punnett square.

Wait, let's correct: Let's say the dark, zebra - striped mussel has genotype \(Aa\) (heterozygous) and the solid light - colored has genotype \(aa\) (homozygous recessive). When we do a cross \(Aa\times aa\), the Punnett square gives:

| | A | a |
|---|---|---|
| a | Aa | aa |
| a | Aa | aa |

So the genotypes of the offspring are \(Aa\) (50%) and \(aa\) (50%). But the dark, zebra - striped shells are associated with the heterozygous genotype (\(Aa\))? Wait, no, the problem says "dark, zebra - striped shells" – maybe the heterozygous is zebra - striped. So the number of offspring with dark, zebra - striped shells (genotype \(Aa\)) would be 50% of the total. Wait, but the options are 6%, 50%, 75%, 100%. Wait, no, maybe the cross is between \(AA\) (zebra - striped) and \(Aa\)? No, the problem says "cross a mussel with dark, zebra - striped shells (genotype \(AW\)) with a mussel with solid light - colored shells (genotype \(aa\))". Wait, maybe the correct cross is between a heterozygous (\(Aa\)) and a homozygous recessive (\(aa\)). The Punnett square for \(Aa\times aa\) is:

Offspring genotypes: \(Aa\) and \(aa\), each with a 50% chance. But if the zebra - striped is the heterozygous genotype (\(Aa\)), then the number of offspring with zebra - striped (dark, zebra - striped) shells would be 50%, which is option B (50%).

Step2: Calculate Probability

The cross is between a heterozygous (\(Aa\)) and a homozygous recessive (\(aa\)) individual. Using the Punnett square method:

- The heterozygous parent (\(Aa\)) produces two types of gametes: \(A\) and \(a\), each with a probability of 50%.
- The homozygous recessive parent (\(aa\)) produces only \(a\) gametes.

When we combine the gametes:

- \(A\) from the first parent and \(a\) from the second parent gives \(Aa\) (50% chance).
- \(a\) from the first parent and \(a\) from the second parent gives \(aa\) (50% chance).

If the dark, zebra - striped shells are associated with the \(Aa\) genotype, then 50% of the offspring will have the same dark, zebra - striped shells.

Answer:

B. 50%