A student attaches a rope to the handle of a chest loaded with a few books, while the other end of the rope is tied to a spring
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Answer:
The children that the couple may have a 25% chance of having cystic fibrosis and 75% chance of being carriers for the CF gene
Explanation:
We know that both parents are in the same situation. Let's start for example with the mother. She has one parent and one sibling with cystic fibrosis (CF). So, supposing that the allele of the gene that causes the disease is <em>"f"</em>, as the disease is autosomal recessive, both, the parent and the sibling have the genotype ff.
Also, knowing that the other parent has a normal phenotype, but the sibling has CF, we deduce that the other parent of the mother must have a <em>Ff</em>.
So, <em>despite the mother's phenotype is normal, her genotype is Ff</em> (see the Punnett square 1) Couple's parents in the attached image). In other words, she is a carrier for the CF gene.
As the parents' situation is analogous, <em>we can deduce that the father also has a genotype Ff </em>and is also a carrier for the CF gene.
Now, if we look at the Punnett square for them (see the square 2) Couple in the attached image), we see that all their children will always have at least one copy of the CF allele, <em>f</em>.
More specifically, their children have 25% chance of having cystic fibrosis and 75% chance of having a normal phenotype but being carriers of the CF gene.
