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earnstyle [38]
3 years ago
8

A cytologist is examining a tissue under an electron microscope. He notices that the endoplasmic reticulum of each cell is extre

mely rough in appearance and he knows that the rough appearance is because of the ribosomes embedded there. He asks why there are so many ribosomes. You respond, _________.A. This tissue exports lipids and is very involved with mRNA which of course is used in protein synthesis.B. This tissue exports proteins to other areas of the body. C. This tissue exports lipids to other areas of the body.D. This tissue exports various nucleic acids, hence the large number of ribosomes present on the endoplasmic reticulum in each of those cells.
Biology
1 answer:
Eddi Din [679]3 years ago
4 0

Answer:

B. This tissue exports proteins to other areas of the body.

Explanation:

The endoplasmic reticulum is a tubular, flattened sac-like structure that forms a continuous structure with the nucleus.

The endoplasmic reticulum may or may not be attached to another organelle called the ribosomes. The ribosome is associated with the translation of mRNA molecules to protein synthesis.

As the protein is synthesized, the proteins enter the endoplasmic reticulum which modifies the proteins and then package them for the transport to their destined site.

Since the endoplasmic reticulum is involved in the process of transport as it modifies the protein molecule therefore is most appropriate option.

Thus, Option-B is correct.

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Mendel's law of independent assortment states that __________ randomly assort their __________ during the formation of haploid c
Shalnov [3]

Answer:

alleles ,genes

Explanation:

Mendel's law of independent assortment states that alleles randomly assort their genes during formation of haploid cells. The paired factors that determine an organism's characteristics segregate separately from each other during sex cell formation.

4 0
3 years ago
10. Which of the following is an example of a prevention activity? A. Mobilizing search and rescue teams B. Requiring identifica
Step2247 [10]

Answer:

Option (B).

Explanation:

Fundamentals of Emergency Management (FEMA) may be defined as the organization that helps in the management of resources. This also helps to deal with the all  humanitarian emergency.

Prevention activities to remove the particular danger are also listed in the FEMA.  The identification for the site access previously might helps to reduce the health related and risk related problems of the individuals working at the site.

Thus, the correct answer is option (B).

5 0
3 years ago
The fact that cats and humans are both classified as mammals provides us with which minimum of information?
Aleks04 [339]

Answer:

a. option :both have mammary glands is the correct answer right no

5 0
2 years ago
Which of the following would be part of an aquatic ecosystem?
puteri [66]
Plankton is part of an aquatic ecosystem
8 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
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