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worty [1.4K]
3 years ago
6

On what basis are enzymes classified and named?

Biology
1 answer:
kenny6666 [7]3 years ago
5 0

Answer: D

Explanation:

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In a swampy marshland, there can be frogs, turtles, snakes, alligators.
victus00 [196]

Answer:

Before discussing more about freshwater swamp forests, we need to know in ... Animals that live in swamps include alligators, amphibians, shellfish, bears and panthers. ... and cypress knobs provide a rich, sheltered habitat for nesting birds, as well as ... All kinds of amphibians (frogs, toads and salamanders) can be found.

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8 0
2 years ago
Which of the following is an example of Genetic Engineering?
crimeas [40]
Answer is B in my opinion
6 0
3 years ago
Which RNA types are involved in translation?
Nezavi [6.7K]
C. Is the answer for sure
4 0
3 years ago
HELP ASAP DUE TONIGHT!!!!
stiv31 [10]

Answer:Over time, carbon-14 decays in predictable ways. And with the help of radiocarbon dating, researchers can use that decay as a kind of clock that allows them to peer into the past and determine absolute dates for everything from wood to food, pollen, poop, and even dead animals and humans.

Explanation:

8 0
3 years ago
In a hardy-weinberg population with two alleles, a and a, that are in equilibrium, the frequency of allele a is 0.2. what is the
Allushta [10]
The Hardy-Weinberg equation is as follows:
( {p + q})^{2}  = 1
{p}^{2}  +2pq +   {q}^{2}  = 1
Where:
(convert all % to decimals)

p= homozygous dominant
q= homozygous recessive
pq= heterozygous

While you did not specify whether the 0.2 frequency was for dominant or recessive, we can still figure out the answer.

Using the 1st equation, we can solve for the other dominant/recessive frequency:

1-0.2=0.8

Meaning that:

p= 0.8 & q=0.2

If the heterozygouz frequency is 2pq, then it becomes a simple "plug & chug" sort of approach.

2(0.8)(0.2)= 2(0.16)= 0.32

So, the heterozygous frequency would be:
0.32

Hope this helps!
8 0
3 years ago
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