Two people(phenotypically normal) have two children. The oldest is normal but youngest has cystic fibrosis.?

Question

What is happening in this cross?

And what is the chance, given in terms of percentage, that the older child is not a carrier of this disease?

Answer 1

Cystic fibrosis is a recessive disease. The oldest child inherited either one dominant allele and one recessive allele or two dominant alleles. So he does not have the disease. The younger child inherited two recessive genes for the disease. He, therefore, has cystic fibrosis.
By the way, both parent are heterozygous (Bb) for the disease and, as such, are carriers. The older child may be a carrier-33% chance.

Answer 2

Both parents must be carriers of the recessive cf gene. They must therefore both have a dominant non-cf gene. The chance of the child being a carrier of cf is 50%, of not being a carrier is 25% and of having cf is 25%. If the child is a carrier it means that it has inherited the cf gene from one parent but not from the other. If it is not a carrier it means the child has inherited the normal gene from both parents. You can get a really simple test done at the GP which just involves the inside of the child's cheek being scraped to see if the child is a carrier. Hope this makes sense, the cystic firosis trust gives out loads of info which explains everything much clearer then I can! Their address is 11 London Road Bromley BR11BY. Possibly contact them for more advice.

Answer 3

Any person with cystic fibrosis is homozygous for a mutant recessive allele. So, the two parents HAVE to both be heterozygotes (Cc) because they are both normal, but had a child with the disease.

Since they are both heterozygous, the expected proportions of offspring will be 3/4 normal, 1/4 with CF. Of the normal offspring, 1/3 will be CC, and 2/3 will be Cc. So the chance that the normal child is not a carrier would be 33% (1/3)

Answer 4

The most likely reason for the outcome is that both parents are carriers of CF. About 1/20 people carry the CF allele, but as it is recessive, heterozygotes are phenotypically normal it can remain 'hidden' in carriers for several generations. You can't rule out the possibility that one parent was a carrier and new mutation in the gamete of the other parent led to him/her passing on the CF gene, but that is a rarer event.

In the obvious case then both parents were of genotype

+/cf and each could produce + (normal) or cf gametes.



Offspring from this cross could be

++ +cf cf+ or cf cf each with chance of 1/4.

As +cf and cf+ are both carrier genotypes (they are identical it's just that in each the cf came from a different parent) then the chance of the unaffected child being a carrier is 50%.

If this is a real situation that affects you, the potential carrier can have their genotype checked using a simple test.

Answer 5

If the youngest has cystic fibrosis, the parents are both carriers. Percentage chance of the oldest being a carrier = 50%.

Answer 6

by using a techniques the main simple reason would be that the two mom and dad are heterozygous carriers of the recessive allele (Cc). One quarter of their babies would be envisioned to inherit the recessive allele from the two mom and dad thsu having the (cc) genotype for cystic fibrosis. lots greater strangely, one or the two alleles ought to be from de novo mutations. One confirm could be a service (Cc) and the different confirm (CC) homozygous. A mutation interior the CFTR gene on chromosome 7 from the homozygous (CC) confirm ought to bring about a (c) recessive allele being surpassed from that confirm with a 50% possibility of mixing with the recessive allele from the heterozygous confirm ensuing interior the recessive genotype (cc). yet another unusual yet documented phenomenon is uniparental disomy (UPD). usual fertilization ends up in a zygote receiving one replica of each and every chromosome from each and every confirm. In uncommon situations of non-disjunction a gamete (egg or sperm) is produced the place the chromosomes did not divide wisely ensuing in the two copies of a given chromosome segregating into the comparable gamete. the different springing up gamete has no copies of the that chromosome (aneuploidy). Now if one confirm has the heterozygous genotype (Cc) and the different has a common homozygous genotype (CC) a homozygous recessive toddler ought to result. If the heterozygous (Cc) confirm has a non-disjunction of chromosome 7 for the period of meiosis (cellular branch generating gametes) a homozygous recessive gamete (cc) can result. If that gamete then fuses with a gamete from the different that's lacking any copies of chromosome 7, then the ensuing zygote would undergo a recessive homozygous genotype (cc).