Question

mendelian genetic outcomes
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date
directions: answer the following questions - show your work

1. a scientist wants to predict the probability of yellow and green pea crosses. he crosses a yellow homozygous dominant pea (yy) with a green homozygous recessive pea (yy). determine the probability of having all yellow homozygous dominant peas.
2. in humans the trait of having freckles (f) is dominant to not having freckles (f). which genotype and phenotype are correctly paired?

a. ff - no freckles b. ff - no freckles c. ff - freckles d. ff - freckles

1. huntingtons disease is a genetic disorder in humans which is controlled by a dominant gene. if one parent is heterozygous for the trait and the other has normal and recessive alleles, what are their chances of having an offspring with the disease?

a. 25% b. 50% c. 75% d. 100%

1. 7. the offspring of two chickens have 1 out of 4 chances of being white, which is a recessive trait. which cross could produce the probability of such offspring?

a. yy x yy b. yy x yy c. yy x yy d. yy x yy

Explanation:

Step1: Analyze question 21

The cross is \( YY\times yy \). The gametes from \( YY \) is \( Y \), and from \( yy \) is \( y \). Using Punnett square:

$$\begin{array}{|c|c|} \hline & y \\ \hline Y & Yy \\ \hline \end{array}$$

All offspring have genotype \( Yy \), so probability of \( YY \) (yellow homozygous dominant) is \( 0 \).

Step2: Analyze question 22

• For \( FF \): Since \( F \) (freckles) is dominant, \( FF \) has freckles.
• For \( Ff \): Heterozygous, shows dominant trait (freckles).
• For \( ff \): Recessive, no freckles. So \( Ff - \) freckles (option c) is correct.

Step3: Analyze question 23

Let \( H \) (Huntington's) be dominant, \( h \) (normal) be recessive. Cross \( Hh\times hh \). Punnett square:

$$\begin{array}{|c|c|} \hline & h \\ \hline H & Hh \\ \hline h & hh \\ \hline \end{array}$$

Probability of \( Hh \) (diseased) is \( \frac{1}{2}=50\% \) (option b).

Step4: Analyze question 24

White (recessive). Let \( Y \) (non - white, dominant), \( y \) (white, recessive). For \( 1/4 \) white (\( yy \)), parents must be \( Yy\times Yy \). Punnett square:

$$\begin{array}{|c|c|c|} \hline & Y & y \\ \hline Y & YY & Yy \\ \hline y & Yy & yy \\ \hline \end{array}$$

Probability of \( yy \) is \( \frac{1}{4} \) (option c).

Answer:

1. \( 0 \)
2. c. \( Ff - \) freckles
3. b. \( 50\% \)
4. c. \( Yy\times Yy \)