Question

i need help on all of them please & thank you!! Unfortunately Punnett Squares are not always for finding out the best parts about parenting. They are often, as you will see in the next few lessons, often used to determine the probability of a disease being passed on to a child. We’ll have you try one and see what that looks like. One such disease is Cystic Fibrosis. Let’s assume both parents are heterozygous for Cystic Fibrosis. That means both parents carry the recessive gene for the disease, but because of the dominant genes they have, they don’t have it. However, they want to know if they try to have a child, what the chances of the child would be for having Cystic Fibrosis?

Answer 1

question 1:

answered previously

question 2:

a= 50% chance tall daughter

b= 50% chance short daughter

question 3:

a= 75% chance of not having cystic fibrosis

b= 25% chance of having cystic fibrosis

question 3:

Step-by-step explanation:

question 2:

the question specified on the offspring being female so when stating the gametes we must include the XX chromosomes for the female and XY chromosome for the male.

let us denote that:

T = tall gene

t = short gene

Father 》 Mother

parental phenotype: Short 》 tall

parental genotype: XtYt 》 XTXt

parental gametes: (Xt) (Yt) 》 (XT) (Xt)

random fertilization:

Xt Yt

XT XTXt XTYt

Xt XtXt XtYt

F1 generation sex: Male 》Female

F1 chromosomes: XY 》 XX

F1 gen phenotype: Tall 》 Tall

F1 gen genotype: Tt, tt 》 Tt, tt

ratio: 2 : 2 or 1 : 1

percentage gender: 50% 》 50%

percentage tall: 50% 》 50%

question 3:

denoting:

N = normal (dominant)

c= cystic fibrosis (recessive)

parental phenotype: Normal 》Normal

parental genotype: Nc 》 Nc

random fertilization:

N c

N NN Nc

c Nc cc

F1 gen phenotype: Normal 》 Cystic fibrosis

F1 gen genotype: NN, Nc, Nc 》 cc

ratio: 3 : 1

percentage: 75% 》 25%

(if you have anymore questions on the topic let me know)