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Hitman42 [59]
4 years ago
10

Most mutations do not become widespread in a population because they lower an individual’s fitness. True or False?

Biology
1 answer:
Elodia [21]4 years ago
4 0

Answer: True

Explanation:

Most of the mutation is not widespread because they cannot be transferred from one generation to another and moreover due to this reason they are not  naturally selected.

These mutations makes the body of individual weaker and leads to the death of the organism.

If the organism dies how will it transfer the traits so is not favored by nature. This is against the fitness of the individual and the organism becomes weak.

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Which statement is true about eukaryotic cells?
mario62 [17]

Answer:

C is the correct option.

3 0
3 years ago
Directions: For each of the following, write either P for pioneer species or C for climax community in the
adoni [48]

Answer:

1. C

2. P

3. P

-

Volcano Erupts

Lava Forms New Land

Animals Arrive

Explanation:

The pioneer species are hardy species which are the first to colonize barren environments or previously biodiverse steady-state ecosystems that have been disrupted, such as by fire. Meanwhile,the climax community, or climatic climax community, is a historic term for a boreal forest community of plants, animals, and fungi which, through the process of ecological succession in the development of vegetation in an area over time, have reached a steady state.

8 0
3 years ago
Which of the following explains why water is an example of a nonmineral resource?
dangina [55]
<span>Minerals are solids.

That is because when water freezes into ice it becomes a mineral, since it's a solid at that time.
</span>
5 0
3 years ago
Read 2 more answers
What are the diference between aerobic and anaerobic cellular respiration?
stich3 [128]
Aerobic respiration uses oxygen to produce ATP. Anaerobic respiration uses inorganic molecules to produce ATP. I hope that answered your question, my friend. :)
5 0
4 years ago
Read 2 more answers
A certain buffer is made by mixing a weak acid HA and its conjugate base A–. When HA is present at a concentration of 0.5 mM and
MAVERICK [17]

Answer:The ratio of the concentrations of HA  and A^- when the buffer has a pH of 7.02 is 0.69

Explanation:

The dissociation constant for formic acid = K_a=1.8\times 10^{-4}

Concentration of HA= 0.5 mM

Concentration of A^-= 0.1 mM

pH = 6.16

First we have to calculate the value of K_a.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[A^-}{[HA]}

Now put all the given values in this expression, we get:

6.16=pK_a+\log (\frac{[0.1]}{0.5})

pK_a=6.86

To calculate the ratio of the concentrations of HA  and A^- when the buffer has a pH of 7.02.

Using Henderson Hesselbach equation :

7.02=6.86+\log \frac{[A^-]}{[HA]}

7.02=6.86+\log \frac{[A^-]}{[HA]}

\frac{[A^-]}{[HA]}=1.44

\frac{[HA]}{[A^-]}=0.69

Thus the ratio of the concentrations of HA  and A^- when the buffer has a pH of 7.02 is 0.69

7 0
3 years ago
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