The only possibility of the daughter being colour blind is where both parents have a recessive gene for colour blindness with the father being colour blind as well. so assuming the recessive allele is Xc, then the genotype of the mother is XCXc, where she has a normal colour vision and the father's genotype is XcY, where he is colour blind.
I'd go with A. can interact to influence a trait, such as eye color
One is through turgidity. this occurs before ground tissue ( collenchyma and sclerenchyma cells) become well developed to give structural support to the plant as it grows bigger. The xylem tissue (composed of rigid tissue) of the young plant render this support and also maintaining osmotic turgidity of the surrounding plant cells.
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
