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Strike441 [17]
3 years ago
5

Any factor that changes shape of an enzyme can affect the enzyme activity which of the following two factors affect an enzyme op

eration the most
Biology
1 answer:
spin [16.1K]3 years ago
3 0

Answer:

Tempature and Ph affect enzyme activity the most

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Fill in the blanks please I’ll mark you brainliest!
Zolol [24]

Answer:

16. a, research

Explanation:

you should form your hypothesis on research because based on what you researched is what will help you try predict the outcome of the experiment

6 0
3 years ago
An organism grew well on TSA plates, a bit slower on MM1 and not at all on MM1 without glucose. The results indicate that organi
Dafna11 [192]

Answer:

Can grow on MM1.

Explanation:

  • Organisms that can only grow in a media having specific nutrients included in it are called a fastidious organism, this implies that these organisms have specific nutrient requirements for their growth.
  • MM 1 represents the minimal media which has basic nutrients present in it, however as the fastidious organisms require some particular or complex nutrients for their growth they can not grow on the minimal media.
  • Therefore, since the organisms that are given in the question is able to grow on the minimal media MM1, this implies that it is not a fastidious organism.
5 0
3 years ago
When a cell is not dividing, the dna is loosely spread throughout the nucleus in a threadlike form called __________?
Vera_Pavlovna [14]
When a cell is not dividing, the DNA is loosely spread throughout the nucleus in a threadlike form called chromatin.
7 0
3 years ago
Read 2 more answers
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
The table provided shows the predator-and-prey relationships in a wetland habitat.
cupoosta [38]

Answer:

mosquitoes

Explanation:

more population means that they will eat more (decrease in mosquito population)

7 0
3 years ago
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