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ryzh [129]
3 years ago
11

Es

Biology
2 answers:
True [87]3 years ago
5 0

Answer:

Ribosomes,cytoplasm ,cell wall and circular DNA could be used to justify that the cell is a prokaryote.

Explanation:

Ribosome  Ribosome is present in both prokaryotic cell and eukaryotic cell.The prokaryotic cell 70S ribosome which contain 2 subunits the larger subunit is 50S and the smaller subunit is 30s

Cell wall  Prokaryotes such as bacterial cell contain cell wall.Cell wall  of bacteria contain various carbohydrates which are crosslinked with peptides resulting in the formation of peptidoglycan.

Cytoplasm The cytoplasm of prokaryotic cell such as bacteria contain ribosomes,chromosome and extracellular genetic material called plasmid.

Circular DNA Almost all prokayotic chromosome contain circular DNA molecule that does not contain any free ends.

Alina [70]3 years ago
3 0

Answer:

Circular DNA

Explanation:

Organisms were classified into a large group called DOMAIN. The domain consists of two main types of organisms: PROKARYOTES AND EUKARYOTES. The basis of this classification is the possession or not of a well defined nucleus.

Prokaryotic organisms contains a single, circular naked genetic material found in a region of the cytoplasm called NUCLEOID. They do not have a nucleus that houses their genetic material (DNA). However, on the other hand, EUKARYOTES possess a linear chromosome that is enclosed by a membrane bound nucleus.

Therefore, although prokaryotes and eukaryotes have many structural distinctions, one can justify an organism being classified as a prokaryote by the possession of a circular genetic material (DNA).

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Rashid [163]

Answer:

See the answer below

Explanation:

<u>From the problem, color is X-linked. Males have just one (XY) chromosome while females have two (XX). Hence, for a recessive trait, males will need only one allele to be affected while females will need two alleles.</u>

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Crossing the two:

                                         X^CY    x    X^cX^c

   X^CX^c                       X^CX^c                     X^cY                    X^cY

<em>normal female     normal female      affected male      affected male</em>

(a) All their sons would be affected for color deficiency. Hence, the probability of them having a color-deficient son is 100% or 1.

(b) None of their daughters is supposed to be affected for color deficiency. Hence the probability of them having a color-deficient daughter is 0.

<em>If the woman gives birth to a color-deficient but otherwise normal daughter and husband sues for a divorce on the ground of adultery, the case will definitely stand up in the court because none of their daughters can be affected for color deficiency based on genes of the man and the woman. </em>

<em>In order to produce an affected daughter, the father will need to supply an affected X chromosome in conjunction with the one that will be supplied by the woman. </em>

<em>In this case, the man does not have an affected chromosome, hence it is not possible for them to have an affected daughter except that the woman slept with another man who happens to have an affected X chromosome.</em>

<em />

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