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Ksivusya [100]
2 years ago
10

In 1973, Stanley Cohen and Herbert Boyer inserted a gene from an African clawed frog into a bacterium. The bacterium then began

producing a protein directed by the code found on the inserted frog gene. The procedure used by Cohen and Boyer is known as
Biology
2 answers:
BaLLatris [955]2 years ago
8 0
It is known as a transgenic organism. This procedure is otherwise called "genetic engineering." Genes of one species can be altered, or qualities can be transplanted starting with one animal types then onto the next. Genetic engineering is made conceivable by recombinant DNA innovation. Living beings that have modified genomes are known as transgenic.
prohojiy [21]2 years ago
7 0

Answer:

The correct answer is recombinant DNA technology.  

Explanation:

In recombinant DNA technology, the formation of the molecules of DNA is done artificially by combining the DNA from distinct sources. In 1973, Cohen and Boyer separated the genes, which codes for rRNA from the African clawed frog DNA and administered it into the E.coli bacteria DNA. At the time of transcription, the bacteria generated frog rRNA, that is, the initial genetically amended organism.  

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An additional gene, gene W, was also examined. A test cross was made between true-breeding EEWW flies and EEWW flies. The result
Debora [2.8K]

This question is incorrect but here is the correct question below;

An additional gene,gene W was also examined. a test cross was made between true breeding EEWW flies and eeww flies. The resulting F₁ generation was then crossed with eeww flies. 100 offspring in the F₂ generation were examined and it was discovered that the E and W genes were not linked.

Which is the correct genotype of the F₂ offspring if the genes were linked and if the genes were not linked?

a) Linked: 50% EeWw and 50% eeww; not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

b) Linked: 25% Eeww, 50% eeWw; not linked:parental genotypes EeWw and eeww.

c) Linked genotypes (EeWw and eeww) and recombinant genotype ( Eeww & eeWw) in the F₂ generation are nearly the same irrespective of their linkage.

d) Linked: mostly with parental genotypes, Eeww and eeWw; unlinked: 25% EeWw and eeww with 75% Eeww and eeWw.

Answer:

a) Linked: 50% EeWw and 50% eeww; not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

Explanation:

a test cross was made between true breeding EEWW flies and eeww flies

If EEWW self crossed, we have the following ( EW, EW, EW, EW)

Also, for eeww, we have ( ew, ew, ew, ew)

                   

                    EW                   EW                     EW                   EW

ew               EWew               EWew               EWew               EWew      

ew               EWew               EWew               EWew               EWew

ew               EWew               EWew               EWew               EWew

ew               EWew               EWew               EWew               EWew

All offspring are  (EWew)

The question goes further by saying "The resulting F₁ generation was then crossed with eeww flies".

And we are asked to find the correct genotype of the F₂ offspring if the genes were linked and if the genes were not linked

∴

To determine  the offsprings of the linked genes we need to go by the definition and understand what linked genes are: Linked genes are genes that are physically close together on the same chromosomes. Effect of recombinantion on linked genes, results in gene swaps which occur in chromosomes that are homologous.

Having said that; If  EWew × eeww

we have;                 EW   &   ew    ×    ew  &    ew

           EW               ew

ew       EeWw          eeww

ew       EeWw          eeww

offspring that

are linked in   ⇒     EeWw    EeWw     &      eeww      eeww

F₂   will be

\frac{1}{2} = 50% of EeWw of the total 100 offspring in the F₂ cross

\frac{1}{2} = 50% of eeww of the total 100 offspring in the F₂ cross

∴ Linked genes =  50% EeWw and 50% eeww.

For unlinked genes; If  EWew × eeww

if rearrangement occurs in EWew  and EWew self crossed, we have ( EW,Ew,eW,ew) as the traits needed for the unlinked gene F₂ crossing.

Also ewew will be (ew, ew, ew, ew).

                       EW                    Ew                    eW                    ew

ew                  EeWw               Eeww                eeWw                eeww

ew                  EeWw               Eeww                eeWw                eeww

ew                  EeWw              Eeww                 eeWw                eeww

ew                  EeWw              Eeww                 eeWw                eeww

We have the following results for the unlinked genes

\frac{1}{4} = EeWw  25% of the total 100 offspring in the F₂ cross

\frac{1}{4} = Eeww   25% of the total 100 offspring in the F₂ cross

\frac{1}{4} = eeWw   25% of the total 100 offspring in the F₂ cross

\frac{1}{4} = eeww    25% of the total 100 offspring in the F₂ cross

∴ not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

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What every friend is cheater from last time i never get trusted friend
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Answer:

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Explanation:

Common mahn!

Ur intro?

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Do predators really kill only the old and sick prey? What evidence is there for this system
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