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telophase of mitosis; interphase; phases of mitosis; anaphase of mitosis; what happens in anaphase; prophase mitosis; what happe

Kitty [74]

The chromosomes are separated by a structure called the mitotic spindle.

A chromosome is a lengthy DNA molecule with component or all of the genetic material of an organism. In maximum chromosomes the very lengthy skinny DNA fibers are lined with packaging proteins; in eukaryotic cells the maximum vital of those proteins are the histones. These proteins, aided by chaperone proteins, bind to and condense the DNA molecule to keep its integrity. These chromosomes show a complicated third-dimensional structure, which performs a large position in transcriptional regulation.

Chromosomes are generally seen below a mild microscope best for the duration of the metaphase of molecular division.

To know about chromosomes click here

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3 0

1 year ago

Which of the following is the primary cause of the rapid rise in loss of biodiversity?

xz_007 [3.2K]

The best and most correct answer among the choices provided by your question is the third choice or letter C.

Human activity is the primary cause of the rapid rise in loss of biodiversity.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

8 0

3 years ago

Read 2 more answers

Jessica is doing an investigation to see how many birds come to the tree in her backyard each morning. She repeats her

Zolol [24]

Answer:

D

Explanation:

I t would be different because she would of only did it once so not really an investigation

8 0

3 years ago

Yellowfin tuna, or Thunnus albacores, are large sport fish that are found in tropical and subtropical waters around the world. T

Ber [7]

The correct answer is 'secondary consumer’. This is because the yellowfin tuna eats herbivorous fish (these are primary consumers - they only eat plants, which are producers) and take refuge from predators. This shows they are not tertiary consumers, as tertiary consumers are at the top of the food chain.

4 0

3 years ago

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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