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PilotLPTM [1.2K]
3 years ago
6

A 10-year-old child diagnosed with acute glomerulonephritis is admitted to the pediatric unit. the nurse should ensure that whic

h action is a part of the child's care?
Biology
1 answer:
Sati [7]3 years ago
3 0
If a 10 year old child is diagnosed with acute glomerulonephritis and admitted to the pediatric unit, the nurse should TAKE THE VITAL SIGNS EVERY 4 HOURS AND OBTAIN THE DAILY WEIGHT OF THE CHILD.

Glomerulonephritis is a disease of kidney where the glomeruli are injured. It is very important to take measure the vital signs such as taking blood pressure of the child because in this disease acute renal failure might occur that hampers the vital functions of the body. Weighing the child everyday is also important because this helps to know the fluid status in the body.
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True or False: An element is a substance that is made up of one type of atom.
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Explanation:

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2 years ago
In a population of rabbits, there are 496 black rabbits and 27 white rabbits. Fur color is determined by a pair of alleles where
Troyanec [42]

Answer:

0.96

Explanation:

It is given that

B is the dominant allele which represents the black color

and b is the recessive allele which represents the white fur.

B being dominant will result into black color fur for genotype "Bb"

Given -

Frequency of black fur allele (p) is 0.8

As per Hardy Weinberg's  first law of equilibrium

p + q = 1\\

Substituting the value of p in above equation, we get -

q = 1-p\\q = 1-0.8\\q= 0.2

q represents the frequency for white fur allele

Frequency of white fur phenotype is

q^2\\= 0.2^2\\= 0.04

Frequency of homozygous black fur phenotype (BB) is

p^2\\= 0.8^2\\= 0.64

As per Hardy Weinberg's second law of equilibrium -

p^2 + q^2 + 2pq = 1\\0.64 + 0.04 + 2pq = 1\\2pq = 1 - 0.68\\2pq = 0.32\\

Combined frequency of homozygous and heterozygous black fur phenotype is

0.64 + 0.32\\= 0.96

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