<span> One is by titration, another is with a meter, and you can also measure it using colorimetric methods.</span>
Is there any multiple choice questions?
B c a is the correct answer i think
B.
E.
F.
They have the possibility to have a child with the disease
1) Data:
dominant allele- A
recessive allele- a
Heterozygotic: Aa (only carries the allele)
Homozygotic recessive: aa (expresses the disease)
Homozygotic dominant: AA (doesn't carry or expresses the disease)
2)The first cross between Jim and his first wife originated a offspring with
<span>cystic fibrosis</span>. That offspring has the genotype aa and for that to happen both parents need to be a least carriers or express the disease. The exercise rules out the second case, which makes Jim a carrier.
3)Now between Jim and Martha:
From the previous step, we already found that jim is a carrier and the exercise also tells us that Martha is one too. So, their genotype is: Aa.
The cross: Aa x Aa
The probability:
1/4 aa = 25%
2/4 =Aa
1/4 - AA
the probability of having a <span>baby with cystic fibrosis is 25%</span>
