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3 years ago

• How removal of starfish from an aquatic environment/ocean will affect the biodiversity?

1 answer:

6 0

Answer: When the starfish had been eliminated from the location as a part of an experiment, the mussel populace swelled and crowded out different species. The biodiversity of the environment turned into notably reduced. Payne's look at confirmed that figuring out and shielding keystone species can assist hold the populace of many different species.

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Please help a girl out with these 2 questions thanks lol

matrenka [14]

Answer: first photo in my opinion would be d or c, but mostly d!

2nd photo would be either a!

hope i helped!

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3 years ago

Which of the following best pairs the leaf structure with its role

Sergio039 [100]

What are the choices ?

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Exposure of zebrafish nuclei to meiotic cytosol resulted in phosphorylation of NEP55 and L68 proteins by cyclin-dependent kinase

Tpy6a [65]

Answer: THEY ARE INVOLVED IN THE DISASSEMBLY OF THE NUCLEAR ENVELOPE

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3 years ago

In plants, chloroplasts are necessary for A. respiration. B. photosynthesis. C. secretion. D. both respiration AND secretion. E.

Ksivusya [100]

Answer:

B- Photosynthesis

Explanation:

Chloroplasts are chlorophyll-containing, eukaryotic cell structures that function in photosynthesis by absorbing energy from sunlight, combining this energy with water and CO2 to convert them to sugars . This cell structure is known as a plastid. The sugars produced, are important for the survival of the plant.

Chloroplasts reproduce on their own, independent of the whole cell because they contain their own DNA. Plant chloroplasts are located in guard cells in plant leaves. Closely linked to these guard cells are tiny pores called stomata, which allow gas exchange required for photosynthesis.

Photosynthesis occurs in two stages:

The light reaction stage
The dark reaction stage

The Light reaction stage takes place in the presence of light. Clorophyll converts light into chemical energy in the form of ATP and NADPH. Both molecules produced, are used in the dark stage to produce sugar.

In the dark reaction stage, the stroma, containing enzymes, facilitates reactions leading to the production of sugars from ATP and NADPH. This process is also called the carbon fixation stage. The sugar produced can be stored in the form of starch for other processes such as respiration.

8 0

4 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 = $\frac{1}{2}c + \frac{1}{2}c$ .........(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 = $\frac{1}{2} A + \frac{1}{2} a$

..................................................................................

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

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

..........................................................................................

$\frac{1}{4}AA + \frac{1}{2}Aa +\frac{1}{4} aa$

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