Answer:
The recombination frequency between two genes exhibits a positive correlation with the distance between them, that is, farther they are, and more will be the chance of recombination. Thus, recombination frequency is used to signify distance among the two genes, for example, 1 percent recombination frequency demonstrates distance of 1 map unit.
Let us consider that the heterozygous female of genotype AaBb can generate four kinds of gametes, that is, AB, Ab, aB and ab. Of these, the two gametes are the outcomes of recombination, or it can be said that 50 percent are recombinants. Thus, it can be concluded that in case of two linked genes, the maximum probable recombination frequency is 50 percent.
This shows that any genes, which are distant than 50 map units will function as unlinked and will function as if they were on distinct chromosomes, and the frequency of recombinant frequency will be 50 percent.
In the given question, it is given that the map distance between the two genes is 80 map units, that is, more than 50 map units. The maximum probable recombinant offspring will be 50 percent of the entire offspring.
A
Explanation:
The genome of prokaryotes has no introns hence their mRNA does not need splicing like in eukaryotic cells. Also, because the genome of prokaryotes is not delimited from the cytoplasm by a nuclear membrane, ribosomes can attach to the elongating mRNA during transcription and begin translation. Therefore translation of mRNA occurs concurrently with transcription which cannot happen with eukaryotic cells.
In the nucleus of eukaryotic cells, transcription results to a nascent mRNA which is spliced into a mature mRNA.The mature mRNA has to travel outside the nucleus to the cytoplasm to be translated by ribosomes.
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