9. lethal dominant achondroplasia (dwarfism) is caused by a dominant gene. a woman and a man with dwarfism marry. if homozygous achondroplasia results in death of embryos, list the genotypes and phenotypes of all potential live - birth offspring. what is the expected ratio of dwarfism to normal offspring? 10. sex - linked the genes for hemophilia are located on the x chromosome. it is a recessive disorder. list the possible genotypes and phenotypes of the children from a man normal for blood clotting and a woman who is a carrier. (hint: you have to keep track of what sex the children are!) extra credit: remember those roan cows from question 2? they also have a second gene for horn vs. hornless cattle. the allele for horns dominates the allele for hornless. if a bull and cow are heterozygous for both genes, what are the probabilities for each possible phenotype?

Question

Accepted Answer

### Problem 9: Lethal Dominant (Achondroplasia)
#### Step 1: Determine Genotypes
- Achondroplasia is caused by a dominant gene ($D$), and homozygous dominant ($DD$) is lethal (embryonic death).
- The woman and man both have dwarfism (achondroplasia) but are not $DD$ (since $DD$ is lethal), so their genotypes must be $Dd$ (heterozygous, as $DD$ is lethal and they are alive with dwarfism).

#### Step 2: Set Up Punnett Square
|       | $D$ | $d$ |
|-------|-------|-------|
| $D$ | $DD$ (lethal, not live) | $Dd$ (dwarf) |
| $d$ | $Dd$ (dwarf) | $dd$ (normal) |

#### Step 3: Analyze Live Offspring
- Total possible offspring genotypes from Punnett square: $DD$ (lethal, 1/4), $Dd$ (2/4), $dd$ (1/4).
- Excluding $DD$ (lethal), the live offspring are $Dd$ (2) and $dd$ (1), so the ratio of dwarfism ($Dd$) to normal ($dd$) among live offspring is $2:1$.

#### Step 4: List Genotypes and Phenotypes of Live Offspring
- **Genotypes**: $Dd$ (2/3 chance) and $dd$ (1/3 chance) (since we exclude the lethal $DD$).
- **Phenotypes**:
  - $Dd$: Dwarfism (achondroplasia)
  - $dd$: Normal (no dwarfism)

### Problem 10: Sex - Linked (Hemophilia)
#### Step 1: Determine Genotypes
- Hemophilia is a recessive X - linked disorder. Let $X^H$ be the normal allele and $X^h$ be the hemophilia allele.
- Man: Normal for blood clotting. Since men have one X and one Y chromosome, his genotype is $X^HY$ (because if he had $X^h$, he would have hemophilia as he has only one X chromosome).
- Woman: Carrier. Carriers of X - linked recessive disorders have the genotype $X^HX^h$ (one normal X, one X with the hemophilia allele).

#### Step 2: Set Up Punnett Square for Sex - Linked Inheritance
We consider the possible gametes:
- Man ($X^HY$) produces gametes $X^H$ and $Y$.
- Woman ($X^HX^h$) produces gametes $X^H$ and $X^h$.

The Punnett square for the combination of these gametes is as follows:

|       | $X^H$ (from man) | $Y$ (from man) |
|-------|-------------------|-----------------|
| $X^H$ (from woman) | $X^HX^H$ (female, normal) | $X^HY$ (male, normal) |
| $X^h$ (from woman) | $X^HX^h$ (female, carrier) | $X^hY$ (male, hemophilia) |

#### Step 3: Analyze Genotypes, Phenotypes, and Sex of Children
- **Female Offspring**:
    - Genotypes: $X^HX^H$ (homozygous normal) and $X^HX^h$ (carrier).
    - Phenotypes: Both $X^HX^H$ and $X^HX^h$ females will have normal blood clotting (since the disorder is recessive and they have at least one normal $X^H$ allele).
- **Male Offspring**:
    - Genotypes: $X^HY$ (normal) and $X^hY$ (has hemophilia).
    - Phenotypes: $X^HY$ males have normal blood clotting, $X^hY$ males have hemophilia.

#### Step 4: Calculate Probabilities for Each Genotype and Phenotype
- **Female Children**:
    - Probability of $X^HX^H$: $\frac{1}{4}$ of all possible offspring (1 out of 4 total squares in the Punnett square), and since half of the offspring are female (2 out of 4), the probability among females is $\frac{1}{2}$ (or 50%). Phenotype: Normal.
    - Probability of $X^HX^h$: $\frac{1}{4}$ of all possible offspring, and among females, it is $\frac{1}{2}$ (50%). Phenotype: Carrier (normal blood clotting, but can pass on the hemophilia allele).
- **Male Children**:
    - Probability of $X^HY$: $\frac{1}{4}$ of all possible offspring, and among males, it is $\frac{1}{2}$ (50%). Phenotype: Normal.
    - Probability of $X^hY$: $\frac{1}{4}$ of all possible offspring, and among males, it is $\frac{1}{2}$ (50%). Phenotype: Hemophilia.

### Extra Credit: (Assuming Question 2 was about Roan Cows with Horns/Hornless, let's assume the following)
#### Step 1: Determine Genotypes (Assuming Roan is $Rr$ for coat color, and Horns ($H$) is dominant over Hornless ($h$), and the bull and cow are heterozygous for both genes, so bull: $RrHh$, cow: $RrHh$)
We will use a dihybrid Punnett square, but since the question is about the second gene (horns/hornless) along with roan (from question 2, assuming roan is a codominant trait for coat color, but here we focus on horns/hornless with the given dominance).
- The allele for horns ($H$) dominates hornless ($h$). Both bull and cow are heterozygous for horns: $Hh$ (and also heterozygous for the roan - related gene, but we focus on horns/hornless here).

#### Step 2: Set Up Punnett Square for Horns/Hornless Gene
|       | $H$ | $h$ |
|-------|-------|-------|
| $H$ | $HH$ (horns) | $Hh$ (horns) |
| $h$ | $Hh$ (horns) | $hh$ (hornless) |

#### Step 3: Calculate Probabilities for Phenotypes
- Probability of horns ($HH$ or $Hh$): $\frac{3}{4}$ (since $HH$ (1/4) + $Hh$ (2/4) = 3/4)
- Probability of hornless ($hh$): $\frac{1}{4}$

(Note: If the roan - related gene also affects the phenotype, and we need to consider the combination, but with the given information about horns/hornless dominance and heterozygosity, the above is the probability for the horns/hornless phenotype. If we consider the roan coat color (assuming roan is $Rr$ with codominance) along with horns/hornless, we would do a dihybrid cross, but the question asks for the probabilities for each possible phenotype related to horns/hornless (and potentially roan, but with the given info about horns dominance, the horns/hornless probabilities are as above).)

### Answers

#### Problem 9
- **Expected ratio of dwarfism to normal (live offspring)**: $2:1$
- **Genotypes of live offspring**: $Dd$ (dwarf, $\frac{2}{3}$ chance) and $dd$ (normal, $\frac{1}{3}$ chance)
- **Phenotypes of live offspring**: Dwarf (achondroplasia) and Normal (no achondroplasia) with a ratio of $2:1$

#### Problem 10
- **Genotypes and Phenotypes of Children**:
    - **Females**:
      - Genotype $X^HX^H$: Phenotype - Normal (no hemophilia), Probability - $\frac{1}{4}$ (of all offspring) or $\frac{1}{2}$ (of female offspring)
      - Genotype $X^HX^h$: Phenotype - Carrier (normal blood clotting, can pass hemophilia), Probability - $\frac{1}{4}$ (of all offspring) or $\frac{1}{2}$ (of female offspring)
    - **Males**:
      - Genotype $X^HY$: Phenotype - Normal (no hemophilia), Probability - $\frac{1}{4}$ (of all offspring) or $\frac{1}{2}$ (of male offspring)
      - Genotype $X^hY$: Phenotype - Hemophilia, Probability - $\frac{1}{4}$ (of all offspring) or $\frac{1}{2}$ (of male offspring)

#### Extra Credit
- **Probability of horns phenotype**: $\frac{3}{4}$
- **Probability of hornless phenotype**: $\frac{1}{4}$ (If considering roan coat color along with horns/hornless, more complex dihybrid cross would be needed, but with the given info on horns/hornless dominance and heterozygosity, these are the probabilities for the horns/hornless phenotype)

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QUESTION IMAGE

Question

Explanation:

Problem 9: Lethal Dominant (Achondroplasia)

Step 1: Determine Genotypes

Step 2: Set Up Punnett Square

Step 3: Analyze Live Offspring

Step 4: List Genotypes and Phenotypes of Live Offspring

Problem 10: Sex - Linked (Hemophilia)

Step 1: Determine Genotypes

Step 2: Set Up Punnett Square for Sex - Linked Inheritance

Step 3: Analyze Genotypes, Phenotypes, and Sex of Children

Step 4: Calculate Probabilities for Each Genotype and Phenotype

Answer:

Problem 9

Problem 10

Extra Credit