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Triss [41]
3 years ago
14

1-. What energy source is directly used to drive the final enzyme in oxygenated, cellular respiration, the ATP synthase?

Biology
1 answer:
Stels [109]3 years ago
6 0

Answer: a. Proton Motive Force

Explanation:

As a process presented in Bacteria, Mitochondria, and Chloroplasts, the chemiosmosis is important to generate ATP from ADP. In this process, we have two main components an electrical potential and a proton concentration gradient, that act in a process called proton-motive force. The ignition starts via the movement of electrons with different energy states via electron carriers.

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Suppose that a farmer wishes to start a you‑pick berry farm, where customers pick their own fruit from the berry bushes. In this
Aleks04 [339]

Answer:

The genotypes of the two original plants the farmer has are as follows:

Plant 1: ss,TT,Aa and Plant 2: Ss,tt,aa.

Explanation:

A genotype is literally the underlying makeup that will produce a result that is visible and measurable, that we know as phenotype. In this particular case, we have a farmer who wishes to establish his farm with a specific type of plant: one that is tall, thornless, and has terminal flowers because they are the best for berry-picking, and that is the business he wishes to establish.

In order to be able to obtain this type of plant, he needs to first count with a starting point, the parent plants from which he will be able to produce the line he wants to establish. But to do that he needs to first know the genetic makeup with which, when he crosses the plants, he will then obtain the phenotypical results he seeks.

The question itself tells us the types of alleles (which are, in simple terms, the holders of the genetic information), that produce the phenotypical characteristics of the plants the farmer has.  So we know that there is one type of bush that is tall, thornless and has terminal flowers, and its alleles are s, t and a. The other bush has short, thorned and axial characteristics, and therefore its alleles are: S, T and A.

When the question then gives us the information about the resulting plants when the originals are selfed, and their characteristics, we are able to tell the genetic makeup, or genotype, of the original two plants the farmer has because we can combine the alleles mentioned given the characteristics (phenotypes) and thus obtain the two parent plants.  

8 0
2 years ago
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The best type of graph to illustrate the change in air temperature over a period of time would be a _________ graph. a. circle b
Pepsi [2]
<span>b. line graph is the answer

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5 0
2 years ago
The vocal ligaments attach between the thyroid cartilage and the
sergejj [24]
Vocal ligament Lies at the free upper edge of the cricothryoid ligament
3 0
3 years ago
Name the process responsible for the formation of glomerular filtrate<br>​
STatiana [176]

Answer: Please refer to:

The process by which glomerular filtration occurs is called renal ultrafiltration. The force of hydrostatic pressure in the glomerulus (the force of pressure exerted from the pressure of the blood vessel itself) is the driving force that pushes filtrate out of the capillaries and into the slits in the nephron.

Explanation:

Not sure but hope it helps.

4 0
2 years ago
Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some
Sphinxa [80]

Answer and Explanation:

<em><u>The number of observed individuals</u></em>:

  • AA 42
  • AG 24
  • GG 21

<u><em>Total number of individuals, N</em></u>= 87 = 42 + 24 + 21

<u><em>Allelic frequencies</em></u>:

  • f(p) = (2 x AA + AG)/ 2 x N

       f (p)= (2 x 42 + 24) /2 x 87

       f (p) = (84 + 24) / 174

       f (p)= 108 / 174

      f (p) = 0.62

  • f (q) = (2 x GG + AG)/2 x N

        f (q) = (2 x 21 + 24 )/2 x 87

        f (q) = (42 + 24)/ 174

        f (q) = 66/174

        f (q) = 0.38

p + q = 1

0.62 + 0.38 = 1

<em><u>The expected genotypic frequency:</u></em>

  • F (AA)= 0.62 ² = 0.3844
  • F (AG) = 2 x A x G = 2 x 0.62 x 0.38 = 0.4712
  • F (GG) = 0.38 ² = 0.1444

AA + AG + GG = 0.3844 + 0.4712 + 0.1444 = 1

<u><em>The number of expected individuals</em></u>:

AA= (0.62)² x 87 = 0.3844 x 87 = 33.44

AG= (0.4712) x 87 = 40.99

GG= (0.38)² x 87 = 12.563

<u><em>Total number of expected individuals</em></u> = 33.44 + 40.99 + 12.563 = 87

<u><em>Chi square</em></u>= sum (O-E)²/E

  • AA= (O-E)² /E

        AA=(42 - 33.44) ² / 33.44

        AA= 2.2

  • AB= (O-E)² /E      

        AB= (24 - 40.99)²/ 40.99

        AB=7.04

  • BB=(O-E)² /E

        BB= (21-12.563)²/12.563

        BB= 5.66

<u><em>Chi square</em></u>= sum ((O-E)²/E) = 2.2 + 7.04 + 5.66 = 14.9

<u><em>Degrees of freedom</em></u> = genotypes - alleles = 3 - 1 = 2

p value less than 0.05

There is enough evidence to reject the nule hypothesis. The genotype frequencies are not in equilibrium.

5 0
3 years ago
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