The answer is d. Hope it help
Answer:

Explanation:
An abiotic factor is a part of an ecosytem, but it is not alive. Some examples include water, rocks, sunlight, and the atmosphere.
On the other hand, a biotic factor is a living part of an ecosystem. Some examples are animals, plants, and fungi.
Let's look at the answer choices.
A creosote plant (choice A) kangaroo rat (choice C) and coyote (choice D) are all living organisms. Therefore, they must be biotic factors.
However, rocky desert soil (choice B) is not living. It's a part of the ecosystem, but since it's not alive it must be <u>abiotic.</u>
Answer. D
Explanation:
After the messenger RNA (mRNA) is formed, it has to go trough different steps before being translated into proteins, also known as maturation of the mRNA. One of the most important steps is splicing, a process that removes the introns (regions of the sequence that do not codify for any particular amino acid sequence or protein). In other words, the splicing process removes sequences that do not generate any protein, leaving only the exons to be translated into protein. However, the genes (within our DNA) will contain exons and introns; Therefore, the gene sequence will have more kilobases compared to mature mRNA as the introns in this macromolecule have been spliced out.
I hope this clarify your question
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Following the Hardy-Weinberg equilibrium theory, the frequency of the heter0zyg0us genotype is 2pq. In the exposed example, 2pq = 0.48.
<h3>
Hardy-Winberg equilibrium</h3>
The Hardy-Weinberg equilibrium theory states that the allelic frequencies in a locus are represented as p and q.
Assuming a diallelic gene,
→ The allelic frequencies are
- p is the frequency of the dominant allele,
- q is the frequency of the recessive allele.
→ The genotypic frequencies after one generation are
- p² (H0m0zyg0us dominant genotypic frequency),
- 2pq (Heter0zyg0us genotypic frequency),
- q² (H0m0zyg0us recessive genotypic frequency).
If a population is in H-W equilibrium, it gets the same allelic and genotypic frequencies generation after generation.
The addition of the allelic frequencies equals 1 ⇒ p + q = 1.
The sum of genotypic frequencies equals 1 ⇒ p² + 2pq + q² = 1
If the allele A has a frequency of 0.6, and the allele B has a frequency of 0.4, then the frequency of the heter0zyg0us genotype is
2pq = 2 x 0.6 x 0.4 =<u> 0.48</u>
You can learn more about the Hardy-Weinberg equilibrium at
brainly.com/question/3406634
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