Cystic fibrosis is regulated by a recessive allele, c. Explain how two normal parents can produce a child with

Answer 1

Two "normal" parents could only produce a child with cystic fibrosis if they were both carriers of the gene, which would make their genotype Cc. If you do a Punnett square, you can see the genotypes of their possible children along with the percent chance of each.
CC- 25% - normal
Cc- 50% - normal
cc- 25% - cystic fibrosis
SO they would have a 25% chance of having a child with cystic fibrosis.

Answer 2

By far the most common reason would be that both parents are heterozygous carriers of the recessive allele (Cc). One quarter of their children would be expected to inherit the recessive allele from both parents thsu having the (cc) genotype for cystic fibrosis.

MUCH more unusually, one or both alleles could be from de novo mutations. One parent may be a carrier (Cc) and the other parent (CC) homozygous. A mutation in the CFTR gene on chromosome 7 from the homozygous (CC) parent could result in a (c) recessive allele being passed from that parent with a 50% chance of combining with the recessive allele from the heterozygous parent resulting in the recessive genotype (cc).

Another unusual but documented phenomenon is uniparental disomy (UPD). Normal fertilization results in a zygote receiving one copy of each chromosome from each parent. In rare cases of non-disjunction a gamete (egg or sperm) is produced where the chromosomes did not divide correctly resulting in both copies of a given chromosome segregating into the same gamete. The other developing gamete has no copies of the that chromosome (aneuploidy).

Now if one parent has the heterozygous genotype (Cc) and the other has a normal homozygous genotype (CC) a homozygous recessive child could result. If the heterozygous (Cc) parent has a non-disjunction of chromosome 7 during meiosis (cell division producing gametes) a homozygous recessive gamete (cc) can result. If that gamete then fuses with a gamete from the other which is lacking any copies of chromosome 7, then the resulting zygote would bear a recessive homozygous genotype (cc).

Answer 3

Recessive Allele

Answer 4

If both parents are heterozygous (Cc), they will both be normal but 1/4 of their children will have the disease (this occurs when both parents contribute the recessive allele to a child). 1/2 the children will be heterozygous carries, and 1/4 of the children will have no remnants of the mutant gene at all.

Answer 5

Both parents would have to be carriers. For someone to have cystic fibrosis, they would have to have alleles cc. Both parents would have to be Cc, and pass c onto the offspring.

Answer 6

C.F is a hereditary autosomal disease the genes of which are located on chromosome no 7.
its symptoms will only develops when a person is homozygous to the disease . it means that a person must carry both recessive alleles cc on his sex chromosome.
Healthy parents will also produce child with c.f
when both the parents are carriers of this disease .
i.e they are both heterozygous.
its like tat:
mother father
Cc Cc
carrier carrier
\ /

offspring1 offspringII

cc CC
with c.f normal
this the transfer of c.f n a child with tis defect.

Answer 7

Look carefully at the top example here.
Substitute c's for b's and it should be clear.

Answer 8

It may be a bit hard but nothing wrong in trying hard!