Final answer:
The possibility for Lauren's child to inherit cystic fibrosis from a CF-affected mother and a carrier father is 50% for being affected and 50% for being a carrier, with no chance for a child to be non-carrier or unaffected. A Punnett square involving one parent with “ff” and another with ‟Ff” genotypes clearly illustrates these probabilities.
Step-by-step explanation:
When examining cystic fibrosis (CF) inheritance, it is important to understand the basics of autosomal recessive disorders. CF is inherited in an autosomal recessive pattern, which means that two copies of the faulty CF gene (“ff”) are needed for a child to have the disease. According to the information provided, Lauren has cystic fibrosis, which means her genotype for the CF gene must be “ff” since she is affected by the disease. Her partner does not have CF, implying they could either be a non-carrier (‟FF”) or a carrier (‟Ff”) without showing symptoms.
Based on the statement that there is a 50% chance their child will have CF, Lauren's partner must be a carrier (‟Ff”). In the genetic cross, we denote Lauren's alleles as “f” and “f” (since she is affected) and her partner's alleles as ‟F” and ‟f” (as a carrier without symptoms). When we draw a Punnett square for their cross (Lauren “ff” x Partner ‟Ff”), we find that there are two possible results: a child with ‟Ff” genotype (carrier) and a child with ‟ff” genotype (affected by CF).
Note: Since Lauren's genotype is “ff” and can only provide a recessive ‟f” allele, and the partner being a carrier also has a ‟f” allele to contribute, any child they have will at least be a carrier. Therefore, the correct probabilities are 50% for the child being a carrier (‟Ff”) and 50% for the child having CF (‟ff”). There is no probability of a child not being a carrier or affected by CF from this couple. Moreover, without additional genetic information, it is not possible to determine the probability of the child having a different genetic condition.