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pochemuha
3 years ago
13

Question 5 (1 point)

Biology
2 answers:
aliina [53]3 years ago
7 0

Answer:

ca

Explanation:

g100num [7]3 years ago
6 0

Answer:

Resource partitioning

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Gina is lactose intolerant but she loves to eat cheesy fries from her local restaurant. What do you suggest Gina does differentl
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Answer:

A

Explanation:

Its the most logical one

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2 years ago
This model shows a human embryo. What can you determine about its development from the model?
sasho [114]

Answer:

It has a postanal tail.

It has pharyngeal arches.

5 0
3 years ago
Which of the following is a primary function of the active site of an enzyme?
Fiesta28 [93]

The primary function of the active site of an enzyme is to catalyze the reaction associated with the enzyme (Option c). It is a fundamental structure in the enzyme.

<h3>What is the active site of an enzyme?</h3>

The active site of the enzyme is It is a fundamental structure in the enzyme that has catalytic activity.

The active site of the enzyme is a site that binds to the substrate to form the enzyme-substrate complex.

The formation of this complex leads to the generation of one or more products of a given chemical reaction.

Learn more about enzymes here:

brainly.com/question/1596855

3 0
2 years ago
Identify the structure of the human heart which is valve between the left atrium and left ventricle of the heart, consisting of
aliya0001 [1]
MMMMIIIIITTTTRAAAALLLL VALVE
8 0
3 years ago
Alkaptoneuria is a recessive genetic disorder found among North American populations at a frequency of about 1 in 500,000 births
Dafna11 [192]

Answer:

The correct answer is - 1/41,493

Explanation:

Let assume the frequency of the two possible same allele genotype (dominant and recessive) in an inbred population is p and q. Then the frequency of heterozygotes (H) is denoted as:

2pq + 2pqF.  ( where F is the inbreeding coefficient).

The frequency of the two different hoozygotes in inbred population can be calculated as:

p2 + pqF and q2 + pqF. (Where p and q are the allele frequency of the dominant and recessive phenotype.

Given: Frequency of Alkaptonuria (q 2) = 1:500, 000

=> q = 1/707

p = 706/707 ( Approx values)

solution:

Inbreeding coefficient (F) = 1/64

Therefore,

Frequency of Alkaptonuria in second cousins= q 2 + pqF

= 1/500, 000 + (706/707 x 1/707) x (1/64)

= 1/500, 000 + 1/45, 248

= 1/41,493 (approx)

5 0
3 years ago
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