Answer:
0.0177
Explanation:
Cystic fibrosis is an autosomal recessive disease, thereby an individual must have both copies of the CFTR mutant alleles to have this disease. The Hardy-Weinberg equilibrium states that p² + 2pq + q² = 1, where p² represents the frequency of the homo-zygous dominant genotype (normal phenotype), q² represents the frequency of the homo-zygous recessive genotype (cystic fibrosis phenotype), and 2pq represents the frequency of the heterozygous genotype (individuals that carry one copy of the CFTR mutant allele). Moreover, under Hardy-Weinberg equilibrium, the sum of the dominant 'p' allele frequency and the recessive 'q' allele frequency is equal to 1. In this case, we can observe that the frequency of the homo-zygous recessive condition for cystic fibrosis (q²) is 1/3200. In consequence, the frequency of the recessive allele for cystic fibrosis can be calculated as follows:
1/3200 = q² (have two CFTR mutant alleles) >>
q = √ (1/3200) = 1/56.57 >>
- Frequency of the CFTR allele q = 1/56.57 = 0.0177
- Frequency of the dominant 'normal' allele p = 1 - q = 1 - 0.0177 = 0.9823
Answer:
Its D. role of carbon dioxide in photosynthesis.
Explanation:
I know its right bc I just got that question.
Answer:
<u>True</u>
Explanation:
In cells, energy is obtained from food through the process of cellular respiration. In mitochondria, this occurs through aerobic respiration- here, cells breakdown sugars like glucose into carbon dioxide and water and energy in the form of ATP or adenosine triphosphate.
In cellular aerobic respiration:
C6H12O6 (glucose) + 6 O2 → 6 CO2 + 6 H2O + ≅38 ATP
Answer:
I believe you can prove a theory more than a hypothesis. A law is definite.
Answer:
They can not live on their own
Explanation:
The main difference between cells and viruses is that viruses need another cell to infect so that they can reproduce. Cells on the other hand can reproduce on their own.
