Genetics Pt. 1?

Question

A man and a woman are both heterozygous for Cystic Fibrosis, an autosomal recessive disease. They have 16 pre-embryos screened for the disease. How many will be homozygous dominat, homozygous recessive and heterozygous. How many will be affected, how many will be carriers, and how many will not be affected?

Answer 1

C - Normal allele (dominant)
c - Cystic fibrosis allele (recessive)

Affected genotype: cc (homozygous recessive)
Carrier genotype: Cc (heterozygous / carriers)
Unaffected genotype: CC (homozygous dominant)

Man: Cc
Woman: Cc

Probability of embryos having:
CC genotype = 0.25
Cc genotype = 0.50
cc genotype = 0.25

Multiply the probability by the whole number (16) of embryos to get the predicted number of embryos with each genotype:
CC = 0.25 x 16 = 4
Cc = 0.50 x 16 = 8
cc = 0.25 x 16 = 4

Answer 2

CF is inherited exactly as Mendel's work with sweet pea color is inherited. Each.

In your example, if Cc is the woman, a carrier, but not having the disease, and Cc is as well the man, again, a carrier, but not having the disease, the options are

CC Cc cC cc
If the small c is the gene for CF, then each embryo has a 25 % chance of being free of any of the disease (CC), one half of the embryos will be carriers, (Cc, cC) and 25% will have the matching genes to create the disease.

Answer 3

Here is a rule of thumb:

If both parents are heterozygous, then all offsprings have 75% chance of inheriting dominant trait and 25% of inheriting recessive trait.

So, since cystic fibrosis is an autosomal recessive trait, then all offspring has 25% chance of inheriting it.

So theoretically, up to 4(25%) can be possibly effected, and up to 12(75%) will not be affected.


I hope you understand. I tried to make it simple. If you have any other questions please email me, good luck

Answer 4

25% will be Homozygous recessive
25% will be Homozygous dominant
50% will be heterozygous

Affected-25% (4)
Carriers-50% (8)
Not Affected-25% (4)