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Svetradugi [14.3K]
3 years ago
6

All 37 trillion cells in the human body came from?

Biology
1 answer:
Aliun [14]3 years ago
4 0

Answer:

all 37 trillion cell in the body originates from a single fertilized egg called a zygote. The zygote divides repeatedly to produce an embryo. These embryonic cells continue to divide, differentiating into all the cell types present in the body of all humans (and other mammals), from a new-born baby to an elderly adult.

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horrorfan [7]

Answer: A

Explanation:

 Because its the main difference between the 2

7 0
2 years ago
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Atomic mass is determined by the number of protons plus the number of
Luda [366]
Atomic mass is determined by the number of protons plus the number of neutrons 
7 0
3 years 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 =<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
Which statement is true for both plant and animals cells?
bulgar [2K]

Answer:

Nuclei and vacuole

Explanation:

But in animal cell vacuole are temporary exist

3 0
2 years ago
Given is a strand of DNA, fill in the corresponding RNA strand and find which amino acids that strand codes for.
prohojiy [21]

Answer:

A-U-G-<u>C-U-A</u>-A-U-A-<u>C-A-G</u>-U-A-G==== RNA

AUG= Start, CUA= Leu, AUA= Iso,  CAG= Glu, UAG=Stop (AMINO ACIDS)

Explanation:

these are right i know for a fact.. had this unit for school

4 0
2 years ago
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