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Nitella [24]
4 years ago
5

People with diabetes must monitor their blood glucose levels. Most do this by checking it themselves, often several times per da

y. It is important for them to maintain blood glucose levels within a normal range. If a diabetic patient determines that his blood glucose level is too low what should he consume to quickly raise his blood glucose level?
Biology
1 answer:
Ket [755]4 years ago
5 0
Penisilin shot, or drink juice, eat an apple, cereal, anything with sugar.
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On the first day after arriving in Miami for spring break vacation, Matt plays out in the sun for six hours. Later that night he
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Explanation:

its c

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3 years ago
In which cycle does the virus to remain dormant?
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Similarities between pencks and davisian cycle of erosion​
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5 0
3 years ago
small group of 100 people decide to isolate themselves from the world and move to a small and remote deserted island. Out of thi
irina [24]

Answer:

Frequency of p = 0.684

Frequency of p = 0.316

Number of individuals with homozygous dominant (AA) = 47

Number of individuals with heterozygous (Aa)= 43

Number of individuals with homozygous recessive (aa) = 10

Explanation:

Out of 100 people, 10 have albino skin (aa)

So, the frequency of homozygous recessive individuals (q^{2}) is \frac{10}{100} = 0.1

Now, q will be

= \sqrt{q^{2} } = \sqrt{0.1} \\= 0.316

As per Hardy Weinberg's equation -

p + q = 1

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

p + 0.316 = 1p = 1 -0.316\\p = 0.684

Now the frequency of homozygous dominant (AA) will be

p^{2} = 0.684^{2} \\= 0.467

Hence, out of 100 people 0.467 * 100 = 46.7 or 47 people are homozygous dominant (AA)

Like wise out of 100 people 0.1 * 100 = 10 people are homozygous recessive (aa)

As per As per Hardy Weinberg's equation-

p^{2} + q^{2} + 2pq = 1\\

Substituting the values in above equation, we get -

0.467 + 0.316 + 2pq = 1\\2pq = 1 -( 0.467+ 0.1)\\2pq = 0.433

So, out of 100 people 0.433 * 100 = 43.3 or 43 people are heterozygous (Aa)

7 0
4 years ago
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