Answer;
D. should there be laws to protect DNA information;
The question that would at least lead to a scientific investigation based on these findings is; should there be laws to protect DNA information.
Explanation;
Human genome project is a scientific research project with the goal of determining the sequence of nucleotide base pairs that make up human DNA, and of identifying and mapping all of the genes. Its main aim was to provide a complete and accurate sequence of the 3 billion DNA base pairs that make up the human genome and find all the estimated genes
The answer would be d.), an increase in the amount of runoff.
The layer of granitic, sedimentary and metamorphic rocks which form the continents and the areas of shallow seabed close to their shores, known as continental shelves.
<h2>Answer:</h2>
<u>The objects that allow humans to access ground water are:</u>
- <u>A spring</u>
- <u>a well drilled into an aquifer
</u>
- <u>a well drilled below the water table</u>
<h2>Explanation:</h2>
Access to ground water can be gained if we dig a well or use any source that can provide us an access below the water table. A water table is the level below which the ground is saturated with water. So above the saturation level we cannot gain access to water therefore we must go below it. A spring springs from ground below water table and the same thing occurs for well or an aquifer if it is below the water table..
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
