If a couple that is both heterozygous carriers of the cystic fibrosis allele have children, the chance that both their first and second child will be carriers 1/4 or 25% is as determined by the crossing.
<h3>Monohybrid crossing</h3>
The cystic fibrosis condition is a recessive one. Let's assume that the allele for cystic fibrosis is a for the affected and A for the unaffected.
A couple that is both heterozygous would have Aa as their genotype:
Aa x Aa
AA Aa Aa aa
The genotypes and phenotypes of the offspring would be as follows:
AA = unaffected
Aa = unaffected
aa = affected
Chances of an offspring being affected = 1/4 or 25%
Chances of an offspring being affected = 3/4 or 75%
Chances of an offspring being a carrier = 1/2 or 50%
Now, let's calculated the probability.
Probability of the first child being a carrier = 1/2
The probability of the second child also being a carrier = 1/2
The probability of both their first and second child being carriers = 1/2 x 1/2
= 1/4 or 25%
Going by the analysis of the monohybrid cross, the probability of both their first and second child being carriers is 1/4 or 25%.
More on monohybrid crosses can be found here: brainly.com/question/1185199
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