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jolli1 [7]
4 years ago
14

Harry is a healthy man who is 30 years of age. using this information, calculate his approximate target heart rate zone for mode

rate-intensity physical activity.
Biology
1 answer:
klemol [59]4 years ago
7 0
Answer: <span>95 to 133 bpm</span>

The equation for maximum heart rate would be:
max HR= 220- age
max HR= 220-30= 190

The equation to find the target for heart beat in medium intensity exercise would be:
target HR= max HR * 50-70%= 190 * 50-70%= <span>95 to 133 bpm</span>
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The correct answer is :

1. increase in cholesterol level leads to more death due to cardiac disease.

2. Hypothesis: "if the cholesterol level is high in an individual, then the risk of death due to cardiac disease is high"

3. Everyone should limit or control their cholesterol level to avoid cardiac disease or it might be lethal.

Explanation:

In this graph, it is represented that there is an increase in the deaths due to cardiac disease with the increase in the cholesterol in the body. The death increase slowly up to 2.5 gL and after that, there is a huge spike in death, which explains that the higher the cholesterol level higher the chances or risk of death due to cardiac disease.

Hypothesis: "if the cholesterol level is high in an individual, then the risk of death due to cardiac disease is high"

On the basis of this one should control its body cholesterol to avoid the chances of cardiac disease as it may lead to death.

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3 years ago
In humans there is a dominant allele (A) for the absence of moles; while the recessive allele (a) results in the presence of mol
Art [367]

Answer:

a. (p + q)^7 = p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7

b. i. 0.75

   ii. 0.000061

   iii. 0.012

   iv. 0.17

c. 0.67

Explanation:

a. The expansion of the binomial (p + q)7 would be such that:

(p + q)^7 = p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7

b. Both couples are heterozygous:

             Aa    x    Aa

         AA   Aa   Aa   aa

Since A is dominant over a,

probability of having mole (aa) = 1/4

probability of not having moles = 3/4

<em>Therefore, the probability of the first child not having moles </em>= 3/4 or 0.75

ii. Let the probability of not having mole = p and the probability of having mole = q. From the binomial expansion:

(p + q)^7 = p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7

<em>Probability that all of the children will have moles </em>= p^0q^7

since p = 3/4 and q = 1/4

p^0q^7 = (3/4)^0(1/4)^7 = 0.000061

iii. <em>Probability that the first two children will have no moles and the last five will have moles</em> = 21p^2q5

                       = 21(3/4)^2(1/4)^5

                         = 0.012

iv. <em>Probability that 4 will have no moles and 3 will have moles out of the 7 children</em> = 35p^4q^3

               = 35(3/4)^4(1/4)^3

                      = 0.17

c. <em>Probability that the child born without moles is a carrier of the a-allele  = probability of heterozygou</em>s.

From the cross in (b), the genotypes of those born without moles are AA and 2Aa. Therefore, the probability of not having moles and be Aa is:

      = 2/3 or 0.67

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