Cystic fibrosis is a genetic disorder in homozygous recessives that causes death during the teenage years. if 9 in 10,000 newborn babies have the disease,
what are the expected frequencies of the dominant (a1) and recessive (a2) alleles according to the hardy-weinberg equation?
(a) f(a1) = 0.9997, f(a2) = 0.0003(b) f(a1) = 0.9800, f(a2) = 0.0200(c) f(a1) = 0.9700, f(a2) = 0.0300(d) f(a1) = 0.9604, f(a2) = 0.0392

f(A1) = 0.9700, f(A2) = 0.0300

Explanation:

The frequency of homozygous recessive genotype of children is  

As per Hardy weinberg's equation, frequency of homozygous recessive genotype is represented by  

Here

Thus, frequency of recessive allele would be  

as per first law of Hardy weinberg

Hence, option C is correct

a neutral ph of 7 and normal organism temperature is needed for a certain enzyme to be effective.  

what would happen to the enzyme if the ph and/or temperature fluctuate greatly from the norm?  

a.the enzyme’s hydrogen bonds will break.  

b.the activation energy will increase.  

c.more product will be formed.

d.the ideal temperature and ph for the enzyme function will decrease.  

the answer is c. it will give up it's valence electron to the other atom.