A question from Helen: Genetics Problem ?????????????
Cystic fibrosis is a serious genetic disease caused by an autosomal recessive allele (C= normal and c= cystic fibrosis). This trait is not rare amongst humans; approximately 1 in 25 people in human populations are carriers (i.e. heterozygous for this allele). A woman and her husband (neither of whom have the disease) are considering having children, but the woman’s brother (there were only two children in the family) has cystic fibrosis. On the other hand, the woman’s parents and her maternal grandparents (i.e. her mother’s parents) did not have the disease.

a. Draw a partial pedigree showing all of the individuals mentioned and give their genotypes as fully as possible (hint: there may be uncertainty for some).

b. What is the33 probability that the first child produced by this couple will have cystic fibrosis?

No 1 answer:

Answer by Elisa
use a punnett square. its not that difficult.

Do you know better? Why not leave your own answer in the comments below!

Technorati Tags: GENETICS, problem

by mcbill

Question posed by Im Juzt Meh: genetics problem (biology)?
Monohybrid Crosses:
1. Cystic fibrosis is a genetic disorder that causes severe respiratory problems. Two normal, healthy people have a child with cystic fibrosis. what are the henotypes of the mother, father, and child? If this couple has several other childeren, what is the likelihood that they will have another child with cystic fibrosis?

Selected answer:

Answer by Alcaligenes
They both are carrying the allele for cystic fibrosis, which means they are both heterozygous for the gene. If you are studying Mendelian genetics, you probably recognize that some genes are recognized by letters. If we represent the parents as heterozygotes for cystic fibrous, the cross would look like this:

Cc x Cc

If only a cc will develop the disease, then it can be concluded that they have a 25% chance of having a baby with cystic fibrosis.

Good luck!!

Whether you agree or disagree, why not leave your own thoughts below.

Technorati Tags: Biology, GENETICS, problem

by cstmweb

A question from Mark J: genetics problem help?
mabel’s sister died of cystic fibrosis as a child. mabel herself does not have the disease and neither do her parents. mabel is pregnant with her first child. if you were a genetic counselor what would you tell her about the probability that her child will have cystic fibrosis.

The best answer:

Answer by Asst Prof
If there is no history of CF in her husband’s family, I would tell her the chance of having a child with CF is 0.

There is a 2/3 chance of Mabel’s being a carrier, since her parents are both heterozygous. But since CF is an autosomal recessive, if her husband does not have the CF allele, it won’t matter.

Agree or disagree? Leave your own thoughts below.

Technorati Tags: GENETICS, help, problem

by Tanya Dawn

Question by Amy P: Another Hardy-Weinberg Problem. Please HELP!!?
After graduation, you and 19 friends build a raft, sailed to a deserted island, and start a new population, totally isolated from the world. Two of your friends carry (heterozygous for) the receive allele , which in homozygotes causes cystic fibrosis.

a. assuming that the frequency of this allele does not chance as the population grows, what will be the instance of cystic fibrosis on the island?

b. calculate how many times grater cystic fibrosis births on the island are vs. the original mainland. The frequency on mainland is .059%.

Best answer:

Answer by Asst Prof
You and your 19 friends have 40 alleles. Since 2 are carriers, there are 2 n alleles and 38 N alleles (where N=a normal allele). So the frequency of n = 2/40 = 0.05 and the frequency of N = 1-0.05 = 0.95. In the future population, the frequency of cystic fibrosis, nn, will be 0.05↑2 = 0.0025 or 0.25%.

If CF on the mainland is 0.059% or 0.00059, then 0.0025/0.00059 = about 4.24 times more CF on the island.

If you know better then please let us know below.

Technorati Tags: Another, HardyWeinberg, help, PLease, problem

by Tanya Dawn

A question asked by Jessica: hardy weinberg principle problem?
In a certain population, 40 out of every 1000 births results in an individual that suffers from cystic fibrosis. What is the allele frequency for the recessive allele that causes cystic fibrosis?

Most detailed answer:

Answer by Asst Prof
If 40/1000 are cfcf, then the frequency of cfcf is 0.04. The frequency of cf is sqrt(0.04) or 0.2.

What do you think? Leave you answer below!

Technorati Tags: hardy, principle, problem, weinberg

A question from MrSkeptiK: Genetics Word Problem?
What are the chances of the following couple producing a child with the autosomal genetic defect cystic fibrosis. Mark and his old wife, Janet, have a child with cystic fibrosis. His new wife, Ellen, had a brother with cystic fibrosis. Mark and Ellen, neither having cystic fibrosis, have a child. What is the chance that their child will have it?

I already asked it but due to varying results I figured another shot wouldn’t do any harm.

Selected answer:

Answer by Kristin
Cystic fibrosis is a recessive trait. I’ll represent the alleles with the letter C. The genotype CC would be two normal alleles, so no illness and no chance of passing it on. The genotype Cc would be one normal and one affected allele, so no illness, but a chance of passing it on. The genotype cc would be two affected alleles, so the person would have the illness.

Because Mark and Janet had a kid with CF, we know that Mark is a carrier (Cc).

Ellen’s brother had CF, which means that both of her parents were carriers (because there is no mention of her parents being affected themselves). Each time her parents reproduced, they had a 25% chance for CC, 50% chance for Cc, and 25% chance for cc. Because Ellen doesn’t have CF, we know that she is not cc. Thus, she has a 1/3 chance to be CC and a 2/3 chance to be Cc.

If Ellen is CC and reproduces with Mark who is Cc, there is a 0% chance that the child will have CF, so we don’t care about that.

If Ellen is Cc and reproduces with Mark, they have a 25% chance of having a kid with CF.

So we multiply Ellen’s chance of being Cc (2/3) by the chance of Mark and Ellen having a cc baby (25%) and we get 0.167, or 16.7%

Whether you agree or disagree, why not leave your own thoughts below.

Technorati Tags: GENETICS, problem, Word

A question from Aurore: Probability problem having to do with cystic fibrosis?
It is estimated that 1 in 25 Americans is a cystic fibrosis carrier. Find the probability that a randomly selected American couple’s child will have cystic fibrosis, assuming that they’re unrelated?

How do I do this? Thanks for your help; I want to understand the process.

The best answer:

Answer by kelby7670
probability of father a carrier is 1/25
probability of mother a carrier is 1/25
probability of both is 1/25 * 1/25 = 1/625 which is th probability of a cf baby.

Provide your own answer to this question below!

Technorati Tags: cystic, fibrosis, having, Probability, problem

by cstmweb

A question from The Waffle: Another genetics problem?
Imagine that you are a genetic counselor, and a couple planning to start a family come to you for information. Charles was married once before, and he and his first wife had a child with cystic fibrosis. The brother of his current wife Elaine died of cystic fibrosis. What is the probability that Charles and Elain will have a baby with cystic fibrosis? (Neither Charles nor Elaine has cf).

Please explain and not just do the problem

Best answer:

Answer by Mr. Biostud
Considering the fact that cystic fibrosis is an autosomal recessive trait we can represent the affected individual’s genotype as aa. The normal is AA and the carrier is Aa. If this is the case then we can calculate the probability of the couple having a child of their with CF.

Since Charlie do have a child with CF and he himself doesn’t have it we can conclude that he is a carrier (genotype Aa). Moreover we can also confidently suggest that Elaine herself is a carrier base on the premise that his brother died of it and she is not affected. Once again her genotype is Aa.

The probability therefore for this couple to have a child with CF is:

Aa x Aa=1/4AA (normal),1/2Aa (carrier),1/4aa (affected)

The probability therefore is 1/4.

Hope this helps…

What do you think? Leave you answer below!

Technorati Tags: Another, GENETICS, problem

by mcbill

Question posed by ham and cheese: autosomal recessive disorder genetics problem?
Cystic fibrosis is an autosomal recessive disorder. A male whose brother has the disease has children with a female whose sister has the disease. It is not known if either the male or the female is a carrier. If the male and female have one child, what is the probability that the child will have cystic fibrosis. My book says the answer is 1/9, how could that be possible? I keep getting 1/16

The best answer:

Answer by Lori G
Each parent has a 2/3 chance of being a carrier (1/3 chance of being normal, 2/3 chance of carrier, and they obviously don’t have cf). Then there is a 1/4 chance of an affected child if they are both carriers… so 1/4 x 2/3 x 2/3 = 4/36 = 1/9.

hope this helps!

Do you know better? Why not leave your own answer in the comments below!

Technorati Tags: autosomal, disorder, GENETICS, problem, Recessive

by mcbill

Question posed by sam: genetics problem with probability?
Imagine that you are a genetic counselor and a couple planning to start a family come to you for information. Charles was married once before and he and his first wife had a child with cystic fibrosis. The brother of his current wife Elaine died of cystic fibrosis. What is the probability that Charles and Elaine will have a baby with cystic fibrosis? (Neither Charles nor Elaine has cystic fibrosis)

The answer is 1/6 but WHY is this the answer???

No 1 answer:

Answer by Rainbow6562005
If I remember my Punnett Squares correctly, Charles is heterozygous for the defective gene because he had to have had it to contribute to his first child’s death.

Elaine’s parents had to be heterozygous also, since each had to contribute the defective gene to the deceased brother. There is no way, with the information given, to determine Elaine’s inheritance. If she’s carrying the defective gene (heterozygous), there’s a 1/4 chance a particular child will have the disease. If she isn’t (homozygous for the normal gene), there’s no chance any child will develop the disease.

Agree or disagree? Leave your own thoughts below.

Technorati Tags: GENETICS, Probability, problem