5. in humans, assume that spotted skin (d) is dominant over non - spotted skin (d) and that wooly hair (h) is dominant over non - wooly hair (h). cross a marriage between a heterozygous spotted, non - wooly man with a heterozygous wooly - haired, non - spotted woman. give genotypic and phenotypic ratios of offspring.
p1 genotype______ p1 gametes ______, ______, ______, ______
p2 genotype______ p2 gametes ______, ______, ______, ______

2. in horses, black is dependent upon a dominant gene (b) and chestnut upon its recessive allele (b). the trotting gait is due to a dominant gene (t) and the pacing gait to its recessive allele (t). a homozygous black pacer is mated to a homozygous chestnut trotter. give the genotypic and phenotypic ratios.
p1 genotype______ p1 gametes ______, ______, ______, ______
p2 genotype______ p2 gametes ______, ______, ______, ______

Question

5. in humans, assume that spotted skin (d) is dominant over non - spotted skin (d) and that wooly hair (h) is dominant over non - wooly hair (h). cross a marriage between a heterozygous spotted, non - wooly man with a heterozygous wooly - haired, non - spotted woman. give genotypic and phenotypic ratios of offspring.
p1 genotype______ p1 gametes ______, ______, ______, ______
p2 genotype______ p2 gametes ______, ______, ______, ______

2. in horses, black is dependent upon a dominant gene (b) and chestnut upon its recessive allele (b). the trotting gait is due to a dominant gene (t) and the pacing gait to its recessive allele (t). a homozygous black pacer is mated to a homozygous chestnut trotter. give the genotypic and phenotypic ratios.
p1 genotype______ p1 gametes ______, ______, ______, ______
p2 genotype______ p2 gametes ______, ______, ______, ______

Accepted Answer

### Problem 5 (Human Traits Cross)
# Explanation:
## Step 1: Determine Parental Genotypes
- Man: Heterozygous spotted ($Dd$), non - wooly ($hh$) → Genotype $Ddhh$
- Woman: Heterozygous wooly - haired ($Hh$), non - spotted ($dd$) → Genotype $ddHh$

## Step 2: Determine Parental Gametes
- Man ($Ddhh$): Possible gametes are formed by independent assortment. For the $D/d$ pair: $D$ or $d$; for $H/h$ pair: only $h$ (since $hh$). So gametes: $Dh$, $Dh$, $dh$, $dh$ (or simplified $Dh$, $dh$ each with 2 copies, but for Punnett square, we can use $Dh$ and $dh$ as the distinct gametes, but actually, when considering all combinations, it's $Dh$, $Dh$, $dh$, $dh$)
- Woman ($ddHh$): For $D/d$ pair: only $d$; for $H/h$ pair: $H$ or $h$. So gametes: $dH$, $dH$, $dh$, $dh$ (or $dH$ and $dh$ each with 2 copies)

## Step 3: Construct Punnett Square
| | $Dh$ | $Dh$ | $dh$ | $dh$ |
|---|---|---|---|---|
| $dH$ | $DdHh$ | $DdHh$ | $ddHh$ | $ddHh$ |
| $dH$ | $DdHh$ | $DdHh$ | $ddHh$ | $ddHh$ |
| $dh$ | $Ddhh$ | $Ddhh$ | $ddhh$ | $ddhh$ |
| $dh$ | $Ddhh$ | $Ddhh$ | $ddhh$ | $ddhh$ |

## Step 4: Count Genotypes and Phenotypes
- **Genotypes**:
    - $DdHh$: Count the number of cells with this genotype. From the Punnett square, there are 8 cells? Wait, no, let's recount. Wait, the Punnett square is 4 (man's gametes) x 4 (woman's gametes)? Wait, no, earlier we considered man's gametes as $Dh$, $Dh$, $dh$, $dh$ (4 gametes) and woman's as $dH$, $dH$, $dh$, $dh$ (4 gametes). So the Punnett square has 16 cells? Wait, no, I made a mistake. The man's genotype is $Ddhh$, so the possible gametes are $Dh$ and $dh$ (because $hh$ can only give $h$, and $Dd$ gives $D$ or $d$). So the man produces 2 types of gametes: $Dh$ and $dh$, each with a probability of 0.5. The woman's genotype is $ddHh$, so she produces $dH$ and $dh$ gametes, each with probability 0.5. So the correct Punnett square is 2x2:
| | $Dh$ | $dh$ |
|---|---|---|
| $dH$ | $DdHh$ | $ddHh$ |
| $dh$ | $Ddhh$ | $ddhh$ |

Now, if we consider the number of each genotype when we have equal numbers of gametes:
- $DdHh$: 1 (from $Dh\times dH$)
- $ddHh$: 1 (from $dh\times dH$)
- $Ddhh$: 1 (from $Dh\times dh$)
- $ddhh$: 1 (from $dh\times dh$)

Wait, no, actually, when the man produces 2 gametes ($Dh$ and $dh$) and the woman produces 2 gametes ($dH$ and $dh$), the Punnett square has 4 cells. But if we consider the actual number of gametes (since $Ddhh$ can produce 2 $Dh$ and 2 $dh$ gametes, and $ddHh$ can produce 2 $dH$ and 2 $dh$ gametes), the Punnett square is 4x4? No, I think the initial mistake was in the number of gametes. Let's do it correctly.

The man's genotype: $Ddhh$. The possible gametes are determined by the law of segregation. For the $D - d$ locus: $D$ or $d$; for the $H - h$ locus: $h$ (since $hh$). So the gametes are $D h$ and $d h$, each with a frequency of 50% (or in terms of number, if we assume 4 gametes, 2 $Dh$ and 2 $dh$).

The woman's genotype: $ddHh$. For the $D - d$ locus: $d$ (since $dd$); for the $H - h$ locus: $H$ or $h$. So gametes are $d H$ and $d h$, each with a frequency of 50% (or 2 $dH$ and 2 $dh$ gametes).

Now, the Punnett square with 4 (man's gametes) x 4 (woman's gametes) is not correct. It should be 2 (man's gamete types) x 2 (woman's gamete types), but with the number of each gamete considered. So the cross is:

| | $Dh$ (2) | $dh$ (2) |
|---|---|---|
| $dH$ (2) | $DdHh$ (4) | $ddHh$ (4) |
| $dh$ (2) | $Ddhh$ (4) | $ddhh$ (4) |

Wait, no, that's overcomplicating. The correct way is to use the principle of independent assortment and calculate the genotypic ratios by considering the probability of each genotype.

The probability of $DdHh$: Probability of $D$ from man ($0.5$) * probability of $d$ from woman ($1$) * probability of $H$ from woman ($0.5$) * probability of $h$ from man ($1$) = $0.5\times1\times0.5\times1 = 0.25$

Probability of $ddHh$: Probability of $d$ from man ($0.5$) * probability of $d$ from woman ($1$) * probability of $H$ from woman ($0.5$) * probability of $h$ from man ($1$) = $0.25$

Probability of $Ddhh$: Probability of $D$ from man ($0.5$) * probability of $d$ from woman ($1$) * probability of $h$ from woman ($0.5$) * probability of $h$ from man ($1$) = $0.25$

Probability of $ddhh$: Probability of $d$ from man ($0.5$) * probability of $d$ from woman ($1$) * probability of $h$ from woman ($0.5$) * probability of $h$ from man ($1$) = $0.25$

So the genotypic ratio is $DdHh: ddHh: Ddhh: ddhh = 1:1:1:1$

- **Phenotypes**:
    - $DdHh$: Spotted ($Dd$) and wooly - haired ($Hh$) → Spotted, wooly - haired
    - $ddHh$: Non - spotted ($dd$) and wooly - haired ($Hh$) → Non - spotted, wooly - haired
    - $Ddhh$: Spotted ($Dd$) and non - wooly ($hh$) → Spotted, non - wooly
    - $ddhh$: Non - spotted ($dd$) and non - wooly ($hh$) → Non - spotted, non - wooly

Since each genotype has a probability of 0.25, the phenotypic ratio is Spotted wooly: Non - spotted wooly: Spotted non - wooly: Non - spotted non - wooly = $1:1:1:1$

### Problem 2 (Horse Traits Cross)
# Explanation:
## Step 1: Determine Parental Genotypes
- P1 (homozygous black pacer): Black is dominant ($B$), pacer is recessive ($tt$), homozygous black ($BB$) → Genotype $BBtt$
- P2 (homozygous chestnut trotter): Chestnut is recessive ($bb$), trotter is dominant ($TT$) → Genotype $bbTT$

## Step 2: Determine Parental Gametes
- P1 ($BBtt$): For $B/b$ pair: only $B$; for $T/t$ pair: only $t$. So gametes: $Bt$, $Bt$, $Bt$, $Bt$ (or just $Bt$ since all are the same)
- P2 ($bbTT$): For $B/b$ pair: only $b$; for $T/t$ pair: only $T$. So gametes: $bT$, $bT$, $bT$, $bT$ (or just $bT$)

## Step 3: Construct Punnett Square
Since P1 produces only $Bt$ gametes and P2 produces only $bT$ gametes, all offspring will have the genotype $BbTt$

## Step 4: Determine Genotypic and Phenotypic Ratios
- **Genotypic Ratio**: All offspring are $BbTt$, so the genotypic ratio is $BbTt: BbTt: \dots$ (but since all are the same, it's $1$ (for $BbTt$) in a ratio where there's only one genotype. So genotypic ratio: $BbTt = 1$ (or more accurately, since all 4x4 Punnett square cells will be $BbTt$, the ratio is $1:0:0:0$ for other genotypes, but since only one genotype is present, the genotypic ratio is $1$ (for $BbTt$)
- **Phenotypic Ratio**: The phenotype of $BbTt$: Black (since $B$ is dominant) and trotter (since $T$ is dominant). So all offspring are black trotters. So phenotypic ratio: Black trotter = 1 (or $1:0$ for other phenotypes)

# Answers:

### Problem 5
- **Genotypic Ratio**: $DdHh: ddHh: Ddhh: ddhh = 1:1:1:1$
- **Phenotypic Ratio**: Spotted wooly: Non - spotted wooly: Spotted non - wooly: Non - spotted non - wooly = $1:1:1:1$

### Problem 2
- **Genotypic Ratio**: $BbTt = 1$ (or in ratio terms, if we consider possible genotypes, since all are $BbTt$, the ratio is $1$ for $BbTt$ and $0$ for others)
- **Phenotypic Ratio**: Black trotter = 1 (all offspring are black trotters)

	\(Dh\)	\(Dh\)	\(dh\)	\(dh\)
\(dH\)	\(DdHh\)	\(DdHh\)	\(ddHh\)	\(ddHh\)
\(dh\)	\(Ddhh\)	\(Ddhh\)	\(ddhh\)	\(ddhh\)
\(dh\)	\(Ddhh\)	\(Ddhh\)	\(ddhh\)	\(ddhh\)

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QUESTION IMAGE

Question

Explanation:

Problem 5 (Human Traits Cross)

Step 1: Determine Parental Genotypes

Step 2: Determine Parental Gametes

Step 3: Construct Punnett Square

Step 4: Count Genotypes and Phenotypes

Step 1: Determine Parental Genotypes

Step 2: Determine Parental Gametes

Step 3: Construct Punnett Square

Step 4: Determine Genotypic and Phenotypic Ratios

Answer:

Problem 5

Problem 2