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Contact [7]
2 years ago
13

How does the longest stage of cell division stated above play a role for growth of plant

Biology
1 answer:
wariber [46]2 years ago
7 0

Answer:

During interphase, the cell undergoes normal growth processes while also preparing for cell division. It is the longest phase of the cell cycle, cell spends approximately 90% of its time in this phase. In order for a cell to move from interphase into the mitotic phase, many internal and external conditions must be met.

Explanation:

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How are primary and secondary succession similar and how are they different?
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Which three steps describe how an animal obtains and uses energy for growth
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Read 2 more answers
Cystic fibrosis is most common in individuals of Northern European descent, affecting 1 in 3200 newborns. Assuming that these al
IgorLugansk [536]

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

4 0
3 years ago
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