Using the Hardy-Weinberg equations, the percentage of carriers of cystic fibrosis is 8.19%.
<h3>What are the Hardy-Weinberg equations?</h3>
The Hardy-Weinberg equations describe the relationship between alleles in a population that is fairly constant.
The Hardy-Weinberg equations are given below as;

Hardy-Weinberg equations
Percentage of carriers = 2pq
Percentage of carriers = 2 * 0.09 * 0.91 * 100 = 8.19%
Therefore, the percentage of carriers of cystic fibrosis is 8.19%.
Learn more about Hardy-Weinberg equations at: brainly.com/question/16039271
#SPJ11
Answer: If the nucleus didn't exist, the cell wouldn't have direction and the nucleolus, which is inside the nucleus, wouldn't be able to produce ribosomes. ... If the cell membrane were gone, the cell would be uprotected. Everything would lead to the death of the cell.
A. The cell would die.
Explanation:
Lysosomes use enzymes to break down waste
The narrative that a physiological need<span>, </span>such as hunger<span> or thirst, </span>creates an aroused tension state<span> (a </span>drive<span>) that motivates an organism to satisfy the </span>need<span>. An unpleasant </span>physiological tension<span>, </span>such<span> as thirst or </span>hunger<span>, that leads to behavior to </span>reduce<span> the </span>tension<span>.</span>
The Texture And Look Of A Microscopic Object