Depends if they have some of the same genetic structures
Answer:
According to the diagram shown, the section of DNA used to make the mRNA strand is known as a gene (option 2).
Explanation:
Gene consists of a DNA fragment that codes for the synthesis of a specific protein that defines the structural or functional trait of a living being.
When DNA is transcribed into mRNA, each section of transcribed DNA is part of a gene. The information travels to ribosomes in the cytoplasm where protein synthesis occurs.
A complete DNA molecule forms a chromosome, which contains the genes with information to define specific traits of a species.
The other options are not correct because:
<em> 1.Carbohydrate is a biomolecule that is not involved in the genetic process.
</em>
<em> 3. Ribosome is where protein synthesis occurs.
</em>
<em> 4. Chromosome is the complete DNA molecule.</em>
Answer:
A or/and D
Explanation:
Environmental science is the study of how organisms interact with their environment so it can be either a or d. I would say a combination of both.
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
