Anti-infectives are used to treat urinary tract infections which include which anatomical structures? (select all that apply.)

1 answer:

4 0

These anatomical structures are the following:
1. The urine itself which serves as an antiseptic, washing potentially harmful bacteria out from the body during normal urination
2. The ureters join into the bladder in a manner designed to prevent urine from backing up into the kidney when the bladder squeezes urine out through the urethra.
3. The prostate gland in men that secretes infection-fighting substance.
4. The immune system defenses and antibacterial substance in the mucous lining of the bladder eliminate many organisms.
5. The vagina of a healthy women, it is colonized by lactobacilli, a beneficial microorganisms that maintain a highly acidic environment (low pH) that is hostile to other bacteria. It also produces hydrogen peroxide, which help eliminate bacteria and reduces the ability of Escherichia coli (E. coli) to adhere to vaginal cells that is the major bacterial culprit in urinary tract infection.

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Kisachek [45]

Q = recessive allele frequency = 0.3, and thus in H-W equilibrium there are ONLY two alleles, q (recessive) and
p (dominant). Therefore all of the p and q present for this gene in a population must account for 100% of this gene's alleles. And 100% = 1.00.
So p, the dominant allele frequency, must be equal to 1 - q --> p = 1 - q
p = 1 - 0.3 = 0.7.
Since heterozygotes are a combination of the p and q, we must again look at the frequencies of each genotype: p + q = 1, then (p+q)^2 = 1^2
So multiplying out (p+q)(p+q) = 1, we get: p^2+2pq+q^2 = 1 (all genotypes), where p^2 = frequency of homozygous dominant individuals, 2pq = frequency of heterozygous individuals, and q^2 = frequency of homozygous recessive individuals.
Therefore if the population is in H-W equilibrium, then the expected frequency of heterozygous individuals = 2pq = 2(0.7)(0.3)
2pq = 2(0.21) = 0.42, or 42% of the population.
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Explanation:

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