Genetics Worksheets and Printables - Free Printable

Problem Analysis and Solution

The worksheet focuses on blood type genetics, which is determined by three alleles: \( I^A \), \( I^B \), and \( i \). The task involves understanding the inheritance patterns and solving monohybrid crosses. Let's break it down step by step.

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#### Section 1: Introduction

1. Blood Type Phenotypes and Genotypes Summary:
- Type A: Genotypes \( I^A I^A \) or \( I^A i \)
- Type B: Genotypes \( I^B I^B \) or \( I^B i \)
- Type AB: Genotype \( I^A I^B \)
- Type O: Genotype \( ii \)

2. Determine Dominance Patterns:
- i. Type AB (\( I^A I^B \)): Codominance. Both \( I^A \) and \( I^B \) are expressed equally.
- ii. Homozygous Type A (\( I^A I^A \)): Complete dominance. \( I^A \) is fully dominant over \( i \).
- iii. Type O (\( ii \)): Recessive phenotype. Neither \( I^A \) nor \( I^B \) is present.
- iv. Heterozygous Type B (\( I^B i \)): Incomplete dominance (not applicable here; this is a heterozygote with one dominant allele).
- v. Heterozygous Type A (\( I^A i \)): Incomplete dominance (not applicable here; this is a heterozygote with one dominant allele).

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#### Section 2: Monohybrid Cross Problems

##### Problem 2.1: Mother (Heterozygous Type B) and Father (Homozygous Type A)

1. Genotypes of Parents:
- Mother (Heterozygous Type B): \( I^B i \)
- Father (Homozygous Type A): \( I^A I^A \)

2. Monohybrid Cross:
- Parental Gametes:
- Mother: \( I^B \) and \( i \)
- Father: \( I^A \) (only one type since homozygous)
- Punnett Square:
```
| | I^A |
|---|-----|
| I^B| I^A I^B |
| i | I^A i |
```

3. Possible F1 Phenotypes:
- \( I^A I^B \): Type AB
- \( I^A i \): Type A
- Therefore, the possible phenotypes are Type AB and Type A.

4. Probabilities of F1 Blood Types:
- Type O (\( ii \)): 0% (impossible since neither parent has two recessive alleles).
- Type AB (\( I^A I^B \)): 50%
- Type B (\( I^B I^B \) or \( I^B i \)): 0% (impossible since father does not contribute \( I^B \)).
- Type A (\( I^A I^A \) or \( I^A i \)): 50%

---

##### Problem 2.2: Mother (Type O) and Father (Type AB)

1. Genotypes of Parents:
- Mother (Type O): \( ii \)
- Father (Type AB): \( I^A I^B \)

2. Monohybrid Cross:
- Parental Gametes:
- Mother: \( i \) (only one type since homozygous)
- Father: \( I^A \) and \( I^B \)
- Punnett Square:
```
| | I^A | I^B |
|---|-----|-----|
| i | I^A i | I^B i |
```

3. F1 Genotypes and Phenotypes:
- \( I^A i \): Type A
- \( I^B i \): Type B
- Therefore, the F1 genotypes are \( I^A i \) and \( I^B i \), and the phenotypes are Type A and Type B.

4. Probabilities of F1 Blood Types:
- Probability of Type O (\( ii \)): 0% (impossible since neither parent contributes two recessive alleles).
- Probability of Type AB (\( I^A I^B \)): 0% (impossible since mother does not contribute any dominant alleles).
- Probability of Type B (\( I^B i \)): 50%
- Probability of Type A (\( I^A i \)): 50%
- Probability of Type B homozygous (\( I^B I^B \)): 0% (impossible since father does not contribute two \( I^B \) alleles).
- Probability of Type A heterozygous (\( I^A i \)): 50%
- Probability of Type B heterozygous (\( I^B i \)): 50%
- Probability of Type A homozygous (\( I^A I^A \)): 0% (impossible since father does not contribute two \( I^A \) alleles).

---

Final Answers:

#### Section 1: Introduction
- i. Type AB: Codominance
- ii. Homozygous Type A: Complete dominance
- iii. Type O: Recessive
- iv. Heterozygous Type B: Incomplete dominance (not applicable)
- v. Heterozygous Type A: Incomplete dominance (not applicable)

#### Section 2: Monohybrid Cross Problems

##### Problem 2.1:
- a. Genotypes of parents: Mother \( I^B i \), Father \( I^A I^A \)
- b. Possible F1 phenotypes: Type AB, Type A
- c. Probabilities:
- Type O: 0%
- Type AB: 50%
- Type B: 0%
- Type A: 50%

##### Problem 2.2:
- a. P1 genotypes: Mother \( ii \), Father \( I^A I^B \)
- b. F1 genotypes: \( I^A i \), \( I^B i \)
- c. F1 phenotypes: Type A, Type B
- d. Probability of Type O: 0%
- e. Probability of Type AB: 0%
- f. Probability of Type B homozygous: 0%
- g. Probability of Type A heterozygous: 50%
- h. Probability of Type B heterozygous: 50%
- i. Probability of Type A homozygous: 0%

---

Final Boxed Answers:

\[
\boxed{
\begin{array}{l}
\text{Section 1:} \\
\text{i. Codominance} \\
\text{ii. Complete dominance} \\
\text{iii. Recessive} \\
\text{iv. Not applicable} \\
\text{v. Not applicable} \\
\\
\text{Section 2:} \\
\text{Problem 2.1:} \\
\text{a. Mother } I^B i, \text{ Father } I^A I^A \\
\text{b. Type AB, Type A} \\
\text{c. Type O: 0%, Type AB: 50%, Type B: 0%, Type A: 50\%} \\
\\
\text{Problem 2.2:} \\
\text{a. Mother } ii, \text{ Father } I^A I^B \\
\text{b. } I^A i, I^B i \\
\text{c. Type A, Type B} \\
\text{d. 0\%} \\
\text{e. 0\%} \\
\text{f. 0\%} \\
\text{g. 50\%} \\
\text{h. 50\%} \\
\text{i. 0\%} \\
\end{array}
}
\]