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Klio2033 [76]
3 years ago
9

Help me please!!!!! 20 POINTS!!!! Plus brainliest!!! Thanks in advance!!

Biology
1 answer:
Colt1911 [192]3 years ago
7 0

Image 3 -

First let's take a look at the two squares close to the electron transport chain. As you can see, the two boxes are connected, which means there is some type of "conversion" occuring. Oxygen goes in and water comes out. (In other words, upper square = oxygen, lower square = water).

Now, the box that is far from the rest corresponds to the mithocondrial matrix of a mithocondrion, since we're talking about .

The two boxes connected to the ATP synthase are aso connected to each other, one going towards the ATP synthase and the other going out. Now, we just have to guide ourselves with logic. It's called "ATP synthase", which means the product will be ATP. So, ADP goes in and ATP comes out.

Image 4 -

I think some ellimination would do good to answer this question. First of all, salmon contains a reasonable amount of B12 and iron and has high amounts of protein, of course. So options B., D. and E. are incorrect.

Nor salmon nor garlic butter contain fiber, so option A. is also incorrect.

Answer:

C.

Image 5 -

The Krebs cycle, the electron transport chain and respiration (aerobic) are closely related, because in anaerobic respiration we obtain oxygen, which functions as an electron acceptor in the electron transport chain, which produces NAD+ that is later used in the Krebs cycle to produce NADH and FADH2 that are used as electron carryiers in the ETC.

Answer:

C.

Hope it helped,

BioTeacher101.

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Alkaptoneuria is a recessive genetic disorder found among North American populations at a frequency of about 1 in 500,000 births
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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:

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

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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)

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