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slava [35]
3 years ago
8

Which is an accurate description of the growth of a natural animal population?

Biology
2 answers:
andre [41]3 years ago
8 0

Answer:

The answer is A

Explanation:

It is limited by environmental factors. A population of organisms cannot survive if it exceeds its carrying capacity. There are also limiting factors, such as predators, disease, etc.

BlackZzzverrR [31]3 years ago
3 0

the answer is A. it’s limited by environmental factors

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What kind of a simple machine is an ax blade?
sweet [91]
The axe is an example of a simple machine, as it is a type of wedge<span>, or dual inclined plane. </span>
4 0
3 years ago
The ecology question above says : A fast swimming strong body fish such as a swimming which is adapted to clear fast and strong
Dafna11 [192]

Correct question:  A fast swimming strong body fish such as a <u>salmon</u> which is adapted to clear fast and strong water current would most likely be found in which aquatic zone? Options in the attached file

Answer:

Source of a river zone

Explanation:

The fishes body is directly correlated with their environment and the pressures they receive from it. Those animals that are used to living in clear waters, with fast currents must be very strong, must have a hydrodynamic torpedo-like body and big and strong caudal fin to propel against the currents. These animals are used to living in water masses such as the source of a river zone.

Rivers belong to lotic environments, characterized by constant moving water.

  • Profundal zone: Sunlight can not reach this zone, so there are no plants. The Water movement is very limited, and in most cases, water is not clear.
  • Littoral zone: Is the transition zone, near the shore, where the light easily reaches the bottom. Usually, there is plenty of vegetation fixated to the bottom.
  • Wetland zone: In general these are calm water, very rich in diversity, and a proper breeding area for a wide variety of fish.
  • Source of a River zone: Rivers flow from higher to lower areas. They originate at the top of the mountains and fastly descend to finally end in a lake or in the sea.

Some species of salmons characterize by living in the sea during sexual maturity, but whenever they need to breed, they go back to the area where they were born. This is in sweet water. They swim in the river until they reach the place where they are going to oviposit. In order to do this, they need to swim against the current. Many times, females die trying to reach this special place.

7 0
3 years ago
Give an example of what would be membrane enclosed organelles in a eukaryotic cell.​
Gennadij [26K]

Answer:

Mitochondria, endoplasmic reticulum, golgi apparatus, or chloroplasts would all be an example.

Explanation:

Unlike prokaryotic cells, eukaryotic cells have a nucleus enclosed in a plasma membrane, as well as the mitochondrias, endoplasmic reticulums, golgi bodies or apparatuses, and chloroplasts.

3 0
3 years ago
Read 2 more answers
2. Dominant trait: cleft chin (C) Mother’s gametes: Cc
andre [41]

.2. Offspring Genotypes will be Cc or cc.

     Offspring phenotypes : Cleft chin or no cleft chin.

    % chance child will have cleft chin: 50%

3.  % chance child will have arched feet: 25%

4.  % chance child will have blonde hair:  50%

5.  % chance child will have normal vision: 25%

 

Explanation:

CASE 1 :

 Dominant trait: cleft chin (C)

    Recessive trait: lacks cleft chin (c)

    Father’s gametes: cc

    Mother’s gametes: Cc

There are two possible combination of Gametes ,

C fom mother and  c from father= Cc

c from mother and c from father = cc

Gametes of Cc Parents=  \frac{1}{2}C + \frac{1}{2} c........(i)

Gametes of cc parents =<u> </u>\frac{1}{2}c + \frac{1}{2}c .........(ii)

Combining (i) and (ii) we get,

\frac{1}{2}  Cc + \frac{1}{2} cc                              

There fore offspring Genotypes will be Cc or cc

Offspring phenotypes :

Genotype Cc then phenotype= Cleft chin

Genotype cc then phenotype = Lacks cleft chin.

percentage chance child will have cleft chin  =\frac{0.5}{1} ×100

Therefore the chance is 50%.

CASE 2 :

Dominant trait: flat feet (A)

Recessive trait: arched feet (a)

Mother’s gametes: Heterozygous (Aa)

Father’s gametes: Heterozygous   (Aa)

There are four possible combination of genotypes are =AA , Aa, Aa and aa

i.e. A from mother, A from father= AA

     A from mother, a from father =Aa

     a from mother, A from Father = Aa

     a from mother, a from father = aa

Gametes of Aa parent =\frac{1}{2} A + \frac{1}{2} a

Gametes of other Aa parent = \frac{1}{2} A + \frac{1}{2} a

                                       <u>..................................................................................</u>

                                              \frac{1}{4} AA + \frac{1}{4} Aa

                                                                           +  \frac{1}{4} Aa +\frac{1}{4} aa

                                   <u>..........................................................................................</u>

                                <u>\frac{1}{4}AA + \frac{1}{2}Aa +\frac{1}{4} aa</u>

Offspring Genotypes will be: AA or Aa or aa

Offsprings phenotype will be:

Genotype AA then phenotype will be Flat feet

Genotype Aa then phenotype will be flat feet

Genotype aa then Phenotype will be arched feet.

Percentage chance child will have arched feet = \frac{0.25}{1} × 100 = 25%

CASE 3:

Dominant trait: Brown hair (B)

Recessive trait: Blonde hair (b)

Mother’s gametes: Homozygous recessive  (bb)

Father’s gametes: Heterozygous  (Bb)

This case is very similar to the case 1 as one parent is homozygous recessive and other parent is heterozygous.

Resulting in  half  Bb and halve bb combination.

Genotypes will be Bb or bb

Phenotypes will be :

Genotype Bb then phenotype Brown hair

Phenotype bb then Phenotype bb.

% chance child will have blonde hair: 50%

CASE 4:

Dominant trait: farsightedness (F)

Recessive trait: normal vision (f)

Mother’s gametes: Heterozygous  (Ff)

Father’s gametes: Heterozygous  (Ff)

This Case is similar to case 2

it will result in one-fourth FF , half Ff and one-fouth ff combination.

Therefore Genotypes will be: FF, Ff and ff

Phenotypes:

Genotype FF  then phenotype farsightedness

Genotype Ff then phenotype  farsightedness

Genotype ff then phenotype normal vision.

% chance child will have normal vision: 25%

 

3 0
3 years ago
Help please i beg i don’t know
Setler79 [48]

Answer:B, D

Explanation:

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