Cystic fibrosis is inherited from parents only if both have the abnormal CF gene. Children of two parents with the CF gene will either be affected with the disease, a carrier but not affected or not have the gene.
-Write an expression for the genitec makeup of children of two parents that are carriers or cystic fibrosis.
-What is the probability that a child will not be affected and not be a carrier?
This looks more like genetics than algebra.
Let N = normal gene and c = cystic fibrosis gene.
Assuming that both parents are carriers — they carry the gene but do not have CF — both of their genotypes would be Nc.
Use a Punnett square to determine probabilities.
http://en.wikipedia.org/wiki/Punnett_square
I hope this helps. Thanks for asking.
To determine the genetic makeup of children of two parents who are carriers or have cystic fibrosis, we can use basic Punnett square analysis.
Let's denote the normal gene as "N" and the abnormal CF gene as "C". Since both parents are carriers, they each have one normal gene (N) and one abnormal CF gene (C).
In a Punnett square, we can combine the genetic material from the parents to determine the possible genetic outcomes for their children.
Punnett square for parents who are carriers (N/C x N/C):
| N | C |
____|________|_______|
N | NN | NC |
____|________|_______|
C | NC | CC |
____|________|_______|
From the Punnett square, we can see the potential genetic makeup of their children:
- NN: Children with two normal genes will not be affected by cystic fibrosis and will not be carriers.
- NC: Children with one normal gene and one abnormal CF gene will be carriers but unaffected by the disease.
- CC: Children with two abnormal CF genes will be affected by cystic fibrosis.
Now, let's calculate the probability that a child will not be affected and not be a carrier.
In this case, we are interested in the probability of obtaining an NN genotype. From the Punnett square, we can see that there are four possible combinations (NN, NC, NC, and CC). Out of these four combinations, only one results in the NN genotype.
Therefore, the probability of a child not being affected and not being a carrier is 1 out of 4, which can be simplified to 1/4 or 0.25 (or 25%).