Eukaryotic viruses can cause microscopic or macroscopic degenerative changes or abnormalities in host cells and in tissues that are distinct from lysis. These are called

prism b but im not for sure

what's the question ?

an individual’s sex (i.e., whether they are a male or female) is determined by the sex chromosomes. most people have two sex chromosomes, one that is inherited from their mother and one that is inherited from their father. typically, females have two x chromosomes (xx) and males have one x chromosome and one y chromosome (xy). conditions caused by changes (“mutations”) in genes located on the x chromosome are considered x-linked.

x-linked recessive inheritance

most x-linked conditions are recessive. this means that in a person with two x chromosomes (most females), both copies of a gene (i.e., one on each x chromosome) must have a change or mutation whereas in a person with one x chromosome (most males), only one copy of a gene must have a mutation. a female with a mutation in one copy of a gene on the x chromosome is said to be a “carrier” for an x-linked condition. a male with a mutation in a gene on the x chromosome is typically affected with the condition. because females have two copies of the x chromosome and males have only one x chromosome, x-linked recessive diseases are more common among males than females. however, x-linked recessive diseases can occur in both males and females.

for x-linked recessive disorders, an unaffected carrier mother who has a mutation in a gene on the x chromosome can transmit either the x chromosome with this mutation or a “normal” x chromosome to her children. if the father is unaffected, none of her daughters will be affected and all of her daughters will be unaffected—since they will inherit at least one normal x chromosome from their father. however, each daughter will have a 50% chance of being an unaffected carrier like her mother and a 50% chance of both x chromosomes being normal.

we solve the problem using the hardy-wineberg equation.

let us start with the homozygous recessive proportion of individuals with sickle cell anaemia,

we have the proportion as q² = 0.25.

∴ q = 0.5 (frequency of hbs in population)

further,

solving for the frequency of normal cell by using the fact that,

p + q = 1

∴ p = 1 - 0.5 = 1 - 0.5 = 0.5

now we can calculate 2pq in p² + 2pq + q², which is the frequency for heterozygous carriers of sickle cell anaemia,

∴ 2pq = 2 x 0.5 x 0.5 = 0.5

and the homozygous normal cell individuals proportion is given by the expression,

∴ p² = 0.5² = 0.25

hence, the carriers for sickle cell anaemia are in proportion, 0.5, or 50%.