2 years ago

Help plsssssss

Biology

1 answer:

Novay_Z [31]2 years ago

8 0

Answer:

A OPTION IS THE CORRECT ANSWER

PLEASE MARK ME AS THE BRAINLIEST

You might be interested in

To keep a species from getting genetic diseases and inbreeding over time, new

Elan Coil [88]

Answer:

Genetic drifts and meiosis

Explanation:

4 0

3 years ago

Which amino acid residues are phosphorylated by kinases?

Jlenok [28]

Serine, threonine, and tyrosine are most commonly phosphorylated in eukaryotes.

4 0

3 years ago

A species that no longer exists in the wild in a specific area but still lives elsewhere is called...

N76 [4]

Answer:

C.) Extirpated :)

7 0

3 years ago

Read 2 more answers

Which statement is true about allele frequencies in a population?

alina1380 [7]

Answer:

the 2nd and 1st statement is correct about frequencies in a population

7 0

3 years ago

Read 2 more answers

Proteus vulgaris has a doubling time of roughly 28 minutes. If an initial population of 500 cells is allowed to grow for 6 hours

romanna [79]

<u>Answer:</u> The final population of Proteus vulgaris after 6 hours is $3.71\times 10^{6}cells$

<u>Explanation:</u>

We are given:

Proteus vulgaris divides and doubles every 28 minutes

Total time given = 6 hours = 360 min (Conversion factor: 1 hr = 60 min)

Number of times Proteus vulgaris doubles in 6 hours = $\frac{360min}{28min}=12.857times$

Calculating the number of bacteria after 6 hours under ideal conditions:

We are given:

Initial population = 500 cells

Number of times it doubles = 12.857 times

Final Proteus vulgaris population = $500\times 2^{12.857}=3709476.8=3.71\times 10^{6}cells$

Hence, the final population of Proteus vulgaris after 6 hours is $3.71\times 10^{6}cells$

8 0

3 years ago