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3 years ago

List the phases which are included in interphase

1 answer:

8 0

G1- the cell duplicates its organelles and cytosolic components

S phase- DNA is replicated and each chromosomes replicates to form two sister chromatids.

G2- enzyme and protein synthesis take place. ATP is produced and centrioles replicate.

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Because martha leads a mostly sedentary lifestyle, the clinician should provide what type of guidance to help her increase her a

Fiesta28 [93]

Answer;
The guidance that would help increase her activity levels would be to walk short distances and slowly increase total activity time.

Explanation;
A sedentary lifestyle involves excessive amount of daily sitting. Be it to watch television, work the computer, or even read, its negative influences include increased anxiety, cardiovascular disease, diabetes, depression, among others.

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3 years ago

Type B blood has type B markers on red blood cells, .......... antibodies in the plasma, and can donate blood to types........

just olya [345]

First select : Anti-B
Second select : B and AB

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2 years ago

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The cork and bark of trees are formed by ________.

Andrej [43]

they are formed by layers of dead cells and is produced by the formation of multiple layers of suberized periderm, cortical and phloem tissue.

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3 years ago

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14. what are the genotypes of the brown and yellow labs who have all black puppies? show all your work using punnett squares.

Varvara68 [4.7K]

There are two different genotypes of brown labs and three different genotypes of yellow labs (eeBB, eeBb, and eebb) (Eebb and EEbb).

Let's examine genotypes, a different table now: As a result, a brown or yellow lab couple can have both brown labs and black or yellow labs.In actuality, neither brown nor chocolate dogs are permitted. Only yellow puppies can ever be born to a yellow lab couple. Even stranger, two brown labs can have yellow or brown puppies, whereas two black labs can have yellow or black puppies. Only yellow labs are capable of independently producing several shades of colour.

Learn more about genotypes by using this link:

brainly.com/question/12116830

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1 year ago

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