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Crazy boy [7]
3 years ago
15

Many different types of mutations can occur within the body. An individual experiences a mutation that changes a base in a tRNA

strand, but during translation the tRNA strand still creates the same protein. Which type of mutation is responsible for the change in the tRNA base?
Biology
1 answer:
chubhunter [2.5K]3 years ago
5 0

Answer:

Substitution

Explanation:

There are 3 types of mutations: deletions, insertions, and substitutions.

A deletion deletes one of the bases (AGTC), which can completely mess up the sequence and create a completely different protein.

An insertion inserts a new base into the strand, which can also completely mess up the sequence and create a completely different protein.

A substitution just changes one of the bases to a different base. This doesn't usually affect what protein is made, but when it does, it only changes one of the amino acids in the sequence, whereas the others change all of the amino acids in the sequence.

If the same protein is still created, then this person only experienced a substitution because it didn't affect the end result of the protein.

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What is the full form of UV rays?​
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Read 2 more answers
We are studying 3 strains of bacteria, with populations p1, p2, p3, in an environment with three food sources, A, B, C. In a day
Nastasia [14]

Answer:

The population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.

Explanation:

From the given information:

For  food source A; we have:

3P₁ + P₂ + 2P₃ = 58    units of food A ---- (1)

For food source B; we have:

2P₁ + 4P₂ + 2P₃ = 70   units of food B  ---- (2)

For food source C; we have:

P₁ + P₂  = 20   units of food C    ----- (3)

From equation (1) and (2); we have:

3P₁ + P₂ + 2P₃ = 58

2P₁ + 4P₂ + 2P₃ = 70

By elimination method

 3P₁ + P₂ + 2P₃ = 58

-

 2P₁ + 4P₂ + 2P₃ = 70

<u>                                       </u>

<u> P₁  -   3P₂   + 0    = - 12    </u>

P₁ = -12 + 3P₂   ---- (4)

Replace, the value of P₁  in (4) into equation (3)

P₁ + P₂  = 20

-12 + 3P₂ + P₂  = 20

4P₂ = 20 + 12

4P₂ = 32

P₂ = 32/4

P₂ = 8

From equation (3) again;

P₁ + P₂  = 20

P₁ + 8 = 20

P₁  = 20 - 8

P₁  = 12

To find P₃;  replace the value of P₁ and P₂ into (1)

3P₁ + P₂ + 2P₃ = 58

3(12) + 8 + 2P₃ = 58

36 + 8 + 2P₃ = 58

2P₃ = 58 - 36 -8

2P₃ = 14

P₃ = 14/2

P₃ =  7

Thus, the population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.

4 0
3 years ago
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