X-linked traits practice worksheet for genetics education.

Let’s go through each problem one by one. I’ll explain how to solve them step by step, then give you the final answers at the end.

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Problem 1: Becker’s Muscular Dystrophy (X-linked recessive)

This is an X-linked recessive disorder. That means:
- The gene is on the X chromosome.
- Males have only one X (from mom), so if they get the bad allele, they show the disease.
- Females have two X chromosomes — they need two bad alleles to show the disease. If they have one bad and one good, they’re carriers but don’t show symptoms.

We use:
- X^B = normal allele (dominant)
- X^b = disease allele (recessive)

Now let’s fill in genotypes and phenotypes:

✔ Female without Becker’s
→ Could be homozygous dominant (X^B X^B) or heterozygous carrier (X^B X^b). But since it says “without Becker’s”, we assume she doesn’t have the disease — so either genotype works, but typically for these charts, if not specified as carrier, we might list both possibilities? Wait — actually, looking at the table structure, it seems they want specific genotypes that match the phenotype.

But note: In X-linked traits, females can be:
- X^B X^B → normal
- X^B X^b → carrier (no disease)
- X^b X^b → has disease

So for “Female without Becker’s” — this includes both X^B X^B and X^B X^b. But the table probably expects us to pick one? Actually, no — let’s read carefully.

Wait — the instruction says: “Fill in the table below to complete the genotype for each individual described.”

And the descriptions are:

1. Female without Becker’s → could be X^B X^B or X^B X^b — but since it doesn’t specify carrier status, maybe we should write both? Hmm — but the table has single cells. Let me think differently.

Actually, in many textbooks, when they say “female without the disease” for X-linked recessive, they often mean non-carrier unless stated otherwise? No — that’s not right. Carriers also don’t have the disease.

Looking ahead — there’s a row for “Females with Becker’s Normal female” — wait, that seems like a typo? Probably meant “Females with Becker’s” and “Normal female” as separate? Or maybe it's "Females with Becker’s" and then "Normal female" is redundant?

Wait — let’s look again:

The rows are:

- Female without Becker’s
- Male without Becker’s
- Male with Becker’s
- Females with Becker’s Normal female ← this looks like a mistake. Probably it’s “Females with Becker’s” and then “Normal female” is not needed? Or maybe it’s “Carrier female”?

Actually, re-reading: “Females with Becker’s Normal female” — that must be a formatting error. Likely it’s supposed to be:

- Females with Becker’s
- Carrier female

Because later in problem 3, it talks about carriers.

I think the intended rows are:

1. Female without Becker’s → likely means homozygous normal: X^B X^B
2. Male without Becker’s → X^B Y
3. Male with Becker’s → X^b Y
4. Females with Becker’s → X^b X^b
5. Carrier female → X^B X^b

Yes, that makes sense. Because “normal female” might be redundant — perhaps it was meant to be “carrier female”.

So let’s assign:

| Individual Description | Genotype | Phenotype |
|-------------------------------|--------------|-------------------------------|
| Female without Becker’s | X^B X^B | Normal |
| Male without Becker’s | X^B Y | Normal |
| Male with Becker’s | X^b Y | Has Becker’s |
| Females with Becker’s | X^b X^b | Has Becker’s |
| Carrier female | X^B X^b | Normal (but carries allele) |

Note: For “Female without Becker’s”, if they meant any female without the disease, it could include carriers — but since there’s a separate row for carrier, I think here “without Becker’s” means not affected AND not carrier — i.e., homozygous normal.

In some contexts, “without” just means doesn’t have the disease — which would include carriers. But given the table structure, and that carrier is listed separately, I believe:

- “Female without Becker’s” = X^B X^B (non-carrier, normal)
- “Carrier female” = X^B X^b (normal phenotype but carries allele)

That fits best.

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Problem 2: Hemophilia A (X-linked recessive)

Man with hemophilia: genotype = X^h Y (since males have one X, and he has the disease, his X must carry the recessive allele)

Woman who is a carrier: genotype = X^H X^h (she has one normal, one disease allele; doesn’t have disease herself)

They marry. What % of kids expected to have RLA? Wait — what is RLA? Looking back...

In the text: “What percentage of their offspring would be expected to have RLA?”

But earlier it says: “A form of hemophilia called 'RLA' (X-linked agammaglobulinemia)” — oh! So RLA is another name for this condition? Or is it a typo?

Actually, reading: “A form of hemophilia called 'RLA' (X-linked agammaglobulinemia)” — that doesn’t make sense because X-linked agammaglobulinemia is NOT hemophilia. Hemophilia A is factor VIII deficiency. Agammaglobulinemia is immune disorder.

Probably a mistake in the worksheet. But regardless — it says RLA is X-linked recessive genetic disorder, same pattern as hemophilia.

So we treat RLA as X-linked recessive, same as hemophilia.

So parents:

Father: X^r Y (has RLA — so his X has recessive allele; let’s use r for RLA allele, R for normal)

Mother: carrier → X^R X^r

Now do Punnett square:

Gametes from father: X^r or Y
Gametes from mother: X^R or X^r

Offspring:

- Daughter gets X from dad and X from mom:
- X^R (mom) + X^r (dad) → X^R X^r → carrier daughter (no disease)
- X^r (mom) + X^r (dad) → X^r X^r → daughter with RLA

- Son gets Y from dad and X from mom:
- X^R (mom) + Y → X^R Y → normal son
- X^r (mom) + Y → X^r Y → son with RLA

So possible children:

1. X^R X^r → carrier female (normal phenotype)
2. X^r X^r → affected female
3. X^R Y → normal male
4. X^r Y → affected male

Each equally likely — 25% each.

Now, question: “What percentage of their offspring would be expected to have RLA?”

Affected individuals:
- Affected female: X^r X^r → 25%
- Affected male: X^r Y → 25%

Total affected = 25% + 25% = 50%

But wait — is that correct? Let’s list all four outcomes:

Probability:

- 25% chance: carrier female (not affected)
- 25% chance: affected female
- 25% chance: normal male
- 25% chance: affected male

So yes, half the children are affected: the affected female and affected male.

Therefore, 50% of offspring expected to have RLA.

But let me double-check: sometimes people forget that females can be affected too — but in this case, since mom is carrier and dad is affected, daughters can inherit two recessive alleles.

Yes — so 50% overall.

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Problem 3: Woman who is carrier for RLA marries man who does not have the disorder.

Same as above? Wait — now it says:

“A woman who is a carrier for RLA marries a man who does not have the disorder.”

So:

Mother: carrier → X^R X^r
Father: does not have disorder → since male, must be X^R Y (normal)

Now Punnett square:

Mother gametes: X^R or X^r
Father gametes: X^R or Y? No — father produces sperm: either X^R or Y? Wait — males produce sperm with either X or Y? No!

Males have XY, so they produce two types of sperm: half with X chromosome, half with Y chromosome.

But in this case, father is normal male → genotype X^R Y

So his sperm:
- 50% carry X^R
- 50% carry Y

Mother’s eggs:
- 50% carry X^R
- 50% carry X^r

Now offspring:

If egg X^R meets sperm X^R → daughter X^R X^R → normal
If egg X^R meets sperm Y → son X^R Y → normal
If egg X^r meets sperm X^R → daughter X^R X^r → carrier (normal)
If egg X^r meets sperm Y → son X^r Y → affected

So four possibilities, each 25%:

1. Normal daughter (X^R X^R)
2. Normal son (X^R Y)
3. Carrier daughter (X^R X^r) — still normal phenotype
4. Affected son (X^r Y)

So who has RLA? Only the last one — affected son.

That’s 1 out of 4 → 25%

But the question is: “what percentage of their sons may be expected to have RLA?”

Ah! It specifies “sons”, not all offspring.

So among sons only:

Sons are those who get Y from father.

From above, sons are:

- From egg X^R + sperm Y → X^R Y → normal son
- From egg X^r + sperm Y → X^r Y → affected son

Each of these is 50% of the sons? Since mother gives X^R or X^r with equal probability, and father always gives Y for sons.

So among sons: 50% normal, 50% affected.

Therefore, 50% of sons expected to have RLA.

Let me confirm:

Total sons: half of all children are sons.

Of those sons, half get X^R from mom → normal
Half get X^r from mom → affected

So yes — 50% of sons have RLA.

Answer: 50%

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Problem 4: Eye color gene on X chromosome. Red eyes dominant, white eyes recessive.

Cross: red-eyed male × white-eyed female

First, define alleles:

Let’s use:
- X^R = red eye (dominant)
- X^r = white eye (recessive)

Red-eyed male: since male, and red is dominant, he must have X^R Y (because if he had X^r Y, he’d have white eyes)

White-eyed female: female, white eyes is recessive, so must be homozygous recessive: X^r X^r

Now cross:

Male: X^R Y → gametes: X^R or Y
Female: X^r X^r → gametes: all X^r

Offspring:

- Daughters: get X^r from mom and X^R from dad → X^R X^r → red eyes (since dominant)
- Sons: get X^r from mom and Y from dad → X^r Y → white eyes

So all daughters have red eyes, all sons have white eyes.

Question: “what percentage of their offspring could be predicted to have white eyes?”

Only the sons have white eyes.

Assuming equal number of sons and daughters, half the offspring are sons → all of them have white eyes.

So 50% of offspring have white eyes.

List:

- 50% daughters: all red-eyed
- 50% sons: all white-eyed

So white-eyed offspring: 50%

Answer: 50%

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Problem 5: Red-green colorblindness — X-linked recessive

Condition usually found in males.

Why?

Because males have only one X chromosome. If that X has the recessive allele for colorblindness, they will express the trait — there’s no second X to mask it.

Females have two X chromosomes. To be colorblind, they need two recessive alleles — one on each X. That’s much less common.

Most females have at least one dominant normal allele, so they are not colorblind — they might be carriers, but don’t show the trait.

So the reason it’s usually found in males is: males have only one X chromosome, so a single recessive allele causes the disorder, while females need two recessive alleles to show it, which is statistically less likely.

Answer: Because males have only one X chromosome, so if they inherit the recessive allele, they express the trait. Females need two copies of the recessive allele to be affected, which is less common.

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Now, compiling final answers:

For Problem 1 table:

I’ll write the genotypes and phenotypes clearly.

For Problems 2–5, numerical or short answers.

Final Answer section will have all answers in order.

Let me summarize:

Problem 1 Table:

| Individual Description | Genotype | Phenotype |
|-------------------------------|--------------|-------------------------------|
| Female without Becker’s | X^B X^B | Normal |
| Male without Becker’s | X^B Y | Normal |
| Male with Becker’s | X^b Y | Has Becker’s |
| Females with Becker’s | X^b X^b | Has Becker’s |
| Carrier female | X^B X^b | Normal (carrier) |

(Note: I assumed “Females with Becker’s Normal female” was a typo and split into “Females with Becker’s” and “Carrier female”)

Problem 2: 50% of offspring expected to have RLA.

Problem 3: 50% of sons expected to have RLA.

Problem 4: 50% of offspring predicted to have white eyes.

Problem 5: Males have only one X chromosome, so a single recessive allele causes the disorder; females need two recessive alleles, which is less common.

Now, for Final Answer, I need to present them clearly.

Since the user didn't specify format, I'll list them as per problem numbers.

Also, for Problem 1, since it's a table, I'll describe it in text for the final answer.

Alternatively, perhaps just state the genotypes.

But to keep it simple, I'll write:

Final Answer:

1. Genotypes and phenotypes:
- Female without Becker’s: Genotype X^B X^B, Phenotype Normal
- Male without Becker’s: Genotype X^B Y, Phenotype Normal
- Male with Becker’s: Genotype X^b Y, Phenotype Has Becker’s
- Females with Becker’s: Genotype X^b X^b, Phenotype Has Becker’s
- Carrier female: Genotype X^B X^b, Phenotype Normal (carrier)

2. 50%

3. 50%

4. 50%

5. Males have only one X chromosome, so they express the trait if they inherit the recessive allele, while females need two recessive alleles to be affected, making it rarer in females.

I think that covers it.

Let me verify calculations once more.

Problem 2: Parents X^r Y (affected male) and X^R X^r (carrier female)

Offspring:
- Daughters: 50% X^R X^r (carrier, normal), 50% X^r X^r (affected)
- Sons: 50% X^R Y (normal), 50% X^r Y (affected)

So affected: half of daughters and half of sons → total 50% affected. Correct.

Problem 3: Carrier female X^R X^r × normal male X^R Y

Sons: get Y from dad, and X from mom → 50% X^R Y (normal), 50% X^r Y (affected) → so 50% of sons affected. Correct.

Problem 4: Red-eyed male X^R Y × white-eyed female X^r X^r

Daughters: all X^R X^r → red eyes
Sons: all X^r Y → white eyes
So 50% of offspring (the sons) have white eyes. Correct.

Problem 5: Standard explanation for X-linked recessive disorders being more common in males. Correct.

All good.

Final Answer:


1.
- Female without Becker’s: Genotype X^B X^B, Phenotype Normal
- Male without Becker’s: Genotype X^B Y, Phenotype Normal
- Male with Becker’s: Genotype X^b Y, Phenotype Has Becker’s
- Females with Becker’s: Genotype X^b X^b, Phenotype Has Becker’s
- Carrier female: Genotype X^B X^b, Phenotype Normal (carrier)

2. 50%

3. 50%

4. 50%

5. Males have only one X chromosome, so a single recessive allele causes the disorder, whereas females require two recessive alleles to be affected, which is less common.