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rosijanka [135]
3 years ago
6

When translation ends, what happens to the new amino acid chain?

Biology
1 answer:
just olya [345]3 years ago
5 0

Answer:

It folds into a protein.

Explanation:

The folding of the protein is determined by the properties of the side groups of the amino acids in the chain. The interaction between these side chains is dependent on their charges with hydrophobic groups tending to be nestled inside of the protein while the hydrophilic occupy much of the outside. The folding is guided by chaperones.

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The antibacterial activity of penicillin is related to the inhibition of a key bacterial enzyme. Explain this mechanism of enzym
serious [3.7K]

Answer:

Penicillin binds and inhibits bacterial enzyme DD-transpeptidase.

Explanation:

Penicillin is made up of 4 rings of β-lactam and acts by binding to DD-transpeptidase of bacteria. The DD-transpeptidase facilitates the cross-linking of the peptidoglycan cell wall of bacteria. This leads to malformation of the protective cell wall of the bacteria and causes it to lyze and die.

8 0
3 years ago
Practicing three-point shots to improve free-throw accuracy contradicts the _________ principle of exercise.
vodomira [7]
Practicing three point shots to improve free throw accuracy contradicts the SPECIFICITY principle of exercise.
The specificity principle of exercise states that, to become an expert in a particular exercise or skill one should perform that exercise or skill. That is, the things that one do during physical activity should be relevant to one desired outcomes, thus, the training regimen must go from general to specific during training.
5 0
3 years ago
Read 2 more answers
How many copies of the 1st chromosome does a human haploid cell contain
Rufina [12.5K]

Answer:

answer from google: ( see explanation )

Explanation: During the first stage of meiosis, the HOMOLOGUES (1-23) are segregated into different cells, resulting in 2 cells that each have 1 copy of each of the 23 chromosomes (still duplicated from DNA replication).

5 0
3 years ago
A large number of disease-free individuals were enrolled in a study
Nuetrik [128]

Answer:

a,Answer: Type I right censoring, censoring time = 60.

b.This is called an Interval censoring,

c.this is called Random censoring because the year in consideration was chosen at random, censoring time = 61.

d.The observations are left truncated respectively at age 30, 40, 50 and 42.

e.L\alpha \frac{S(60)}{S(30)} *\frac{S(52)-S(55)}{S(40)} *\frac{S(61)}{S(42)} *\frac{S(55)}{S(50)}

Explanation:

a) A healthy individual, enrolled in the study at age 30, never developed

breast cancer during the study.

Answer: Type I censoring, censoring time = 60.

(b) A healthy individual, enrolled in the study at age 40, was diagnosed

with breast cancer at the fifth exam after enrollment (i.e., the disease

started sometime between 12 and 15 years after enrollment).

THis is called an Interval censoring,

if the clinical exams is conducted every three years , then 3*4=12, and 3*5=15

add this to 40 years respectively, we have

censoring time = (52, 55]

c) A healthy individual, enrolled in the study at age 50, died from a

cause unrelated to the disease (i.e., not diagnosed with breast cancer

at any time during the study) at age 61.

this is called Random censoring because the year in consideration was chosen at random, censoring time = 61.

(d) An individual, enrolled in the study at age 42, moved away from

the community at age 55 and was never diagnosed with breast cancer

during the period of observation.

Random censoring, censoring time = 55.

The observations are left truncated respectively at age 30, 40, 50 and 42.

e.  Confining your attention to the four individuals described above,

write down the likelihood for this portion of the study.

L\alpha \frac{S(60)}{S(30)} *\frac{S(52)-S(55)}{S(40)} *\frac{S(61)}{S(42)} *\frac{S(55)}{S(50)}  

the above represent the likelihood

3 0
3 years ago
If a species has 35 percent adenine in its dna, what is the percentage of the other three bases?
Basile [38]

Answer:

The percentage of other bases are: Thymine = 35%, guanine = 15% and cytosine = 15%.

Explanation:

There are four bases in DNA molecules, the bases are adenine, thymine, guanine and cytosine. The quantity of adenine and thymine are always equal, so also, the quantity of guanine and cytosine are always equal. Based on the information given in the question, Adenine = 35%, this implies that thymine too is 35%. The two bases added together give 70% , leaving only 30% for guanine and cytosine. Cytosine and guanine will share this equally and so each one of them will have 15%.

7 0
3 years ago
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