Answer:
The genotype of a woman who is heterozygous for cystic fibrosis Cc.
The genotype of a man who has cystic fibrosis is cc.
The probability that their children will have the disease is 1/2 while the probability that their children will not have the disease is 1/2.
Explanation: If C represents the normal trait and c represents the trait for cystic fibrosis; and if the normal trait is dominant over the trait for cystic fibrosis which is recessive, therefore a woman who is heterozygous for cystic fibrosis will have the genotype Cc while a man who has cystic fibrosis will have the genotype of cc.
In a cross between the man (cc)and the woman (Cc), fifty percent of the children have a genotype (cc), an indication that they have cystic fibrosis while fifty percent of the children have the genotype (Cc) which is heterozygous for cystic fibrosis, an indication that they do not have cystic fibrosis because the trait for cystic fibrosis is recessive.
See the punnett square attached
Given:
The solution decreased by ounces every hour for 5 hours.
Remaining solution at the end of the experiment = ounces.
To find:
The initial amount of solution.
Solution:
Let the initial amount of the solution be x.
The solution decreased by ounces every hour for 5 hours.
Total decreased amount
Initial amount = Remining amount + Decreased amount of solution.
Therefore, the initial amount of solution is 18 ounces.
