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sergiy2304 [10]
3 years ago
10

why does prolonged exposure to UV light from the sun or tanning beds increase the risk of getting skin cancer

Biology
1 answer:
Ratling [72]3 years ago
8 0
In short, because the UV rays damage skin cells, which increases the risk of a mutation possibly leading to skin cancer.

Hope this helps.
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Which of the following statements describe how technology and research changed the understanding of genetics? Select all that ap
Goshia [24]

What are the options?

6 0
3 years ago
Which food is at a temperature that allows bacteria to grow well ? A) a roast at 125•F (52•c) B) hamburgers at 165•f (74•c) C) p
Lilit [14]

The correct answer is A) A roast at 125°F (52°C)

Explanation:

In general terms, bacteria thrive at warm temperatures; this means bacterial growth is lower at extremely hot/cold temperatures, but it is higher at warm or medium temperatures. Indeed, the ideal temperature for bacteria to develop and reproduce is between 4° C and 60°C. This implies from the options given the roast at 52°C represents an ideal temperature for the growth of bacteria. Also, other options include temperatures above 60°C, and therefore do not allow bacteria to grow well.

4 0
3 years ago
Which of the following statements best provides evidence of the big bang theory?
Oxana [17]

Answer: D. Galaxies show a red shift, indicating they are moving away from each other.

Explanation:

Galaxies seem to show a red shift which indicates that the galaxies are moving away from each other. This provides evidence for the Big Bang theory because the theory posits galaxies will keep moving away from each other.

This is because when an explosion happens, matter moves away from the point where the explosion happened. If the Big Bang happened therefore, galaxies will move away from the source which is why galaxies are moving away from each other.

6 0
3 years ago
__________ are waxy substances that circulate in the blood.
artcher [175]
Cholesterol is a waxy substance found in blood.
6 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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