Question

1. two heterozygous black pigs are mated. show the genotypes, phenotypes, and probabilities of the offspring.

what are the genotypes of the parents? _______________

genotypes | phenotypes
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1. in holstein cattle, the spotting of the coat is recessive to the solid coat color. what types of offspring might be produced by a cross between two spotted animals?

what are the genotypes of the parents? _______________

genotypes | phenotypes
--- | ---

1. in peas, long stem \a\ is dominant over short stem, \a\. give the expected genotypes, phenotypes and the probabilities for the following crosses:

a) homozygous long x short

genotypes | phenotypes
--- | ---

b) heterozygous long x short

genotypes | phenotypes
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Explanation:

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Problem 4

Step1: Identify parent genotypes

Heterozygous black pigs: $Bb$ and $Bb$

Step2: Fill Punnett square

$$\begin{array}{c|cc} & B & b \\ \hline B & BB & Bb \\ b & Bb & bb \\ \end{array}$$

Step3: Calculate genotype probabilities

• $BB$: $\frac{1}{4}$ (25%), $Bb$: $\frac{2}{4}=\frac{1}{2}$ (50%), $bb$: $\frac{1}{4}$ (25%)

Step4: Assign phenotypes

Black (dominant $B$): $BB, Bb$; Non-black (recessive $bb$): $bb$
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Problem 5

Step1: Identify parent genotypes

Spotted (recessive, let allele = $s$): $ss$ and $ss$

Step2: Fill Punnett square

$$\begin{array}{c|cc} & s & s \\ \hline s & ss & ss \\ s & ss & ss \\ \end{array}$$

Step3: Calculate probabilities

All offspring: $ss$ (100%)

Step4: Assign phenotypes

Spotted (only $ss$ possible)
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Problem 6a

Step1: Identify parent genotypes

Homozygous long ($AA$) × short ($aa$)

Step2: Fill Punnett square

$$\begin{array}{c|cc} & A & A \\ \hline a & Aa & Aa \\ a & Aa & Aa \\ \end{array}$$

Step3: Calculate probabilities

All offspring: $Aa$ (100%)

Step4: Assign phenotypes

Long stem (dominant $A$)
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Problem 6b

Step1: Identify parent genotypes

Heterozygous long ($Aa$) × short ($aa$)

Step2: Fill Punnett square

$$\begin{array}{c|cc} & A & a \\ \hline a & Aa & aa \\ a & Aa & aa \\ \end{array}$$

Step3: Calculate probabilities

• $Aa$: $\frac{2}{4}=\frac{1}{2}$ (50%), $aa$: $\frac{2}{4}=\frac{1}{2}$ (50%)

Step4: Assign phenotypes

Long stem ($Aa$), short stem ($aa$)

Answer:

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Problem 4

Parent genotypes: $Bb$ and $Bb$

Genotypes	Phenotypes	Probability
$Bb$	Black	50% ($\frac{1}{2}$)
$bb$	Non-black	25% ($\frac{1}{4}$)

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Problem 5

Parent genotypes: $ss$ and $ss$

Genotypes	Phenotypes	Probability

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Problem 6a

Genotypes	Phenotypes	Probability

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Problem 6b

Genotypes	Phenotypes	Probability
$aa$	Short stem	50% ($\frac{1}{2}$)