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scoray [572]
3 years ago
15

In normal cells, stathmin is inactivated by phosphorylation at the start of M phase. Phosphatases remove these phosphates as the

cell transitions from M phase to G1. What enzyme is likely to be responsible for phosphorylating stathmin during M phase
Biology
1 answer:
Sloan [31]3 years ago
5 0

Answer:

The enzyme that is responsible for this is the cyclin dependent kinase subunit of mpf

Explanation:

This enzyme is also known as cyclin dependent kinase subunit, CDKs. they are protein kinases and can function as cell cycle regulators. They have the power to modify different protein substrates that are useful for the progression of cell cycle. CDKs are inhibitors hence stathmin is inactivated by phosphorylation. They help to control cell divisions. DNA should not be damaged for cell to move from G₁ to the S phase.

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4 years ago
A woman is heterozygous for blue eyes. If she were to have child with a man who is homozygous recessive for brown eyes, what is
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3 years ago
Extracts of the rosy periwinkle, Catharanthus roseus, have provided medicine with vincristine and vinblastine, drugs now availab
lions [1.4K]

Answer:

Biodiversity conservation

Explanation:

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6 0
3 years ago
Okay need help on this thing ughhh ..............Online Project:
Igoryamba
So basically two things:
1) you're going to have to flip the coins (or fake numbers) for the experimental trials.
2) for the theoretical, there is 1/2 chance for heads or tails with each toss, so you'd expect that out of 10 tosses, 5 heads, 5 tails. out of 100 tosses- 50 heads, 50 tails.
When tossing 2 coins- 1/2×1/2 = 1/4 (25%) chance that 2 heads, 2 tails, or 1 heads & 1 tails. Deviation value comes from after you done your flipping and recorded your data. So if on 100 flips you actually got 50 and 50 (rarely us that exact ;), the deviation from the expected of 50/50 would be 0.00. If however you flipped 100 heads or 100 tails (impossible), then the deviation value would be 1.00.
|(100-50)| ÷ 50 = 50÷50 = 1.00
So usually you may have data like: 47/53 or something a little off than 50/50, making deviation |(47-50)| ÷ 50 = 3÷50 = 0.06.
Now the number of flips is important for the outcome! So if a coin toss if 10 times had 4 heads, 6 tails, the deviation value would be:
|(4-5)| ÷ 5 = 1÷5 = 0.20

So increasing the # flips DECREASES the deviation value!!
Whether it's from 10 to 100, or from 100 to 200. Look at my example of how the 10-flip deviation of 0.20 decreased to 0.06 with 100-flip
5 0
3 years ago
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