B-4.7 - Genetics Worksheet - Inheritance Patterns - Free Printable

Let’s solve each question step by step with clear explanations.

---

1. Cross between R and R’ (RR, R’R’, R’R)

Given:
- RR = Red
- R’R’ = White
- R’R = Pink

This is a classic example of incomplete dominance.

✔ Type of Inheritance: Incomplete Dominance

In incomplete dominance, the heterozygous genotype (R’R) shows a blended or intermediate phenotype — here, pink — which is neither red nor white.

✔ Genotype: The cross is between two heterozygotes? Wait — looking at the Punnett square:

The parents are labeled R and R’ — but the square has 4 boxes filled with:
Top row: R R
Bottom row: R’ R’

So the cross is R R × R’ R’ → all offspring are R R’

But wait — that doesn’t make sense for a 2x2 grid unless it’s actually R R’ × R R’ — because if both parents are homozygous (RR and R’R’), you’d get all R R’ — but then the Punnett square would show only one genotype.

Looking again: the labels on the sides are “R” and “R’”, implying gametes from each parent. So likely, both parents are heterozygous R R’, producing gametes R and R’.

Then the Punnett square would be:

```
| R | R'
-----------------
R | RR | R R'
R' | R R' | R'R'
```

So offspring genotypes: RR, R R’, R R’, R’R’

Phenotypes: Red, Pink, Pink, White

✔ Genotype (offspring): RR, R R’, R’R’
✔ Phenotype (offspring): Red, Pink, White

But since the question asks “what type of inheritance is it?” — the key is the intermediate phenotype (pink) — so Incomplete Dominance.

---

Answer 1:
- Type of Inheritance: Incomplete Dominance
- Genotype: RR, R’R, R’R’ (or R R’ — same thing)
- Phenotype: Red, Pink, White

---

2. Cross between B and W (BB, WW, BW)

Given:
- BB = Black
- WW = White
- BW = Speckled

Here, the heterozygote (BW) shows both traits simultaneously — speckled = black + white spots.

This is Codominance.

✔ Type of Inheritance: Codominance

In codominance, both alleles are fully expressed in the heterozygote — not blended, but both visible (e.g., speckled coat).

Assuming the cross is B W × B W (heterozygotes), then offspring:

```
| B | W
-----------------
B | BB | BW
W | BW | WW
```

✔ Genotype: BB, BW, WW
✔ Phenotype: Black, Speckled, White

---

Answer 2:
- Type of Inheritance: Codominance
- Genotype: BB, BW, WW
- Phenotype: Black, Speckled, White

---

3. Hemophilia Punnett Square (X-linked recessive)

Parents:
- Mother: X^H X^h (carrier — because she has X^H and X^h)
- Father: X^h Y (affected male — hemophilia is X-linked recessive)

Wait — let’s read carefully:

The Punnett square shows:

Mother’s gametes: X^H and X^h
Father’s gametes: X^h and Y

So mother is X^H X^h (carrier)
Father is X^h Y (has hemophilia)

Offspring:

```
| X^H | X^h
----------------------------
X^h | X^H X^h | X^h X^h
Y | X^H Y | X^h Y
```

So 4 possible children:
- X^H X^h → female carrier
- X^h X^h → female with hemophilia
- X^H Y → male, normal
- X^h Y → male, hemophiliac

Now answer the questions:

a. Mother’s genotype? → X^H X^h
b. Father’s phenotype? → Has hemophilia (since he has X^h Y)
c. Is it possible for daughters to have hemophilia? → Yes — if they inherit X^h from both parents → X^h X^h
d. Is it possible for daughters to be carriers? → Yes — X^H X^h
e. Is it possible for sons to have hemophilia? → Yes — if they inherit X^h from mother → X^h Y
f. Is it possible for sons to be carriers? → No — males have only one X chromosome; they either have the disease or don’t. They cannot be “carriers” like females can.
g. Chance of having a child with hemophilia? → Out of 4 offspring:
- X^h X^h → hemophiliac daughter
- X^h Y → hemophiliac son
→ 2 out of 4 = 50%
h. Type of inheritance? → X-linked recessive

---

Answer 3:
a. X^H X^h
b. Has hemophilia
c. Yes
d. Yes
e. Yes
f. No
g. 50%
h. X-linked recessive

---

4. Pedigree Analysis

Looking at the pedigree:

Generation I:
- Female (shaded = affected) × Male (unshaded = unaffected)

Generation II:
- 4 children: 2 males (unshaded), 2 females (one shaded, one unshaded)

Generation III:
- One of the unshaded males from II marries an unshaded female → has 2 children: 1 male (unshaded), 1 female (shaded)

Key observations:

- Trait appears in females.
- Affected females have unaffected fathers → this rules out X-linked dominant (because if father is unaffected, he gives normal X to daughters — so daughters shouldn’t be affected).
- Also, affected females have unaffected mothers? Wait — in generation I, mother is affected, father is unaffected → daughters can be affected or unaffected.
- In generation III: an unaffected father and unaffected mother have an affected daughter.

That’s critical!

If both parents are unaffected and they have an affected child, the trait must be recessive.

Also, since it affects females and males, and appears in daughters of unaffected parents, it’s autosomal recessive.

Why not X-linked recessive?

In X-linked recessive, for a daughter to be affected, she must inherit mutant X from both parents. That means her father must be affected (since he gives his only X to daughters). But in generation III, the father is unaffected — yet the daughter is affected → impossible for X-linked recessive.

Therefore, it must be autosomal recessive.

✔ Type of Inheritance: Autosomal Recessive

---

Answer 4:
- Autosomal Recessive

---

## ✔ Final Answers Summary:

1.
- Type of Inheritance: Incomplete Dominance
- Genotype: RR, R’R, R’R’
- Phenotype: Red, Pink, White

2.
- Type of Inheritance: Codominance
- Genotype: BB, BW, WW
- Phenotype: Black, Speckled, White

3.
a. X^H X^h
b. Has hemophilia
c. Yes
d. Yes
e. Yes
f. No
g. 50%
h. X-linked recessive

4.
- Autosomal Recessive

---

Let me know if you’d like diagrams or further clarification!