Question

mendelian genetic outcomes
name
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

When crossing \(YY\) (yellow homozygous dominant) with \(yy\) (green homozygous recessive), using the Punnett - square method. The gametes from \(YY\) are all \(Y\), and gametes from \(yy\) are all \(y\). The cross \(YY\times yy\) gives \(Yy\) offspring. The probability of getting \(YY\) (homozygous dominant) is \(0\) since all offspring are \(Yy\) (heterozygous).

Step2: Analyze question 22

For a trait where \(F\) (freckles) is dominant over \(f\) (no freckles):

• For \(FF\), since \(F\) is dominant, the phenotype is freckles.
• For \(Ff\), because \(F\) is dominant, the phenotype is freckles.
• For \(ff\), since there is no dominant allele, the phenotype is no freckles.

Step3: Analyze question 23

Let \(H\) be the dominant allele (Huntington's disease) and \(h\) be the recessive allele (normal). One parent is \(Hh\) (heterozygous) and the other is \(hh\) (homozygous recessive). Using the Punnett - square:
The gametes from \(Hh\) are \(H\) and \(h\), and from \(hh\) are \(h\) and \(h\). The cross \(Hh\times hh\) gives \(Hh:hh = 1:1\). The probability of having an offspring with the disease (\(Hh\)) is \(\frac{1}{2}=50\%\).

Step4: Analyze question 24

If white is a recessive trait (let's say \(yy\) is white). Using the Punnett - square:

• For \(YY\times YY\), all offspring are \(YY\) (non - white).
• For \(YY\times Yy\), offspring are \(YY:Yy = 1:1\) (non - white).
• For \(Yy\times Yy\), \(YY:Yy:yy=1:2:1\). The probability of \(yy\) (white) is \(\frac{1}{4}\).
• For \(yy\times yy\), all offspring are \(yy\) (white, probability \(1\)).

Answer:

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