Answer:
Considering that homeostasis is restored in the patient, his blood pH range would return to normal levels (7.35-7.45), and his hydrogen ion concentration in the blood would normalize. The effect of normalizing the body by getting rid of excess hydrogen ions is achieved by concentrating these ions into the urine for expulsion, therefore increasing the pH levels of urine.
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Explanation:
Acidosis is the condition wherein excessive acid build-up within the body causes the blood pH to become lower than normal (normal pH range 7.35-7.45). This may be due to an excessive loss of bicarbonate in the blood, also known as metabolic acidosis, or due to an impairment in the elimination of carbon dioxide in the blood from poor lung function, also known as respiratory acidosis. The body's natural response to acidosis is to increase the breathing rate to eliminate carbon dioxide in the blood, restoring the natural pH of the body.
In people with diabetes mellitus type I, the lack of insulin causes cells to breakdown fat aside from glucose as an energy source. This process produces ketones as a metabolic by-product for energy but also causes the body to be acidic. This is known as diabetic ketoacidosis.
The rate of evolution is the means of
Darwinism
The answer is A It protects the seed until it matures.
hope this helps
A gemstone is a piece of hard mineral that is considered exquisite in terms of physical properties and valuable because of its rarity or usefulness. In this case, the properties classify diamond as a gemstone is its <span>color and profitability. Diamonds are very expensive because of its rarity</span>
Hardy-Weinberg Equation (HW) states that following certain biological tenets or requirements, the total frequency of all homozygous dominant alleles (p) and the total frequency of all homozygous recessive alleles (q) for a gene, account for the total # of alleles for that gene in that HW population, which is 100% or 1.00 as a decimel. So in short: p + q = 1, and additionally (p+q)^2 = 1^2, or 1
So (p+q)(p+q) algebraically works out to p^2 + 2pq + q^2 = 1, where p^2 = genotype frequency of homozygous dominant individuals, 2pq = genotype frequency of heterozygous individuals, and q^2 = genotype frequency of homozygous recessive individuals.
The problem states that Ptotal = 150 individuals, H frequency (p) = 0.2, and h frequency (q) = 0.8.
So homozygous dominant individuals (HH) = p^2 = (0.2)^2 = 0.04 or 4% of 150 --> 6 people
Heterozygous individuals (Hh) = 2pq = 2(0.2)(0.8) = 0.32 or 32% of 150
--> 48 people
And homozygous recessive individuals (hh) = q^2 = (0.8)^2 = 0.64 = 64% of 150 --> 96 people
Hope that helps you to understand how to solve these types of population genetics problems!
