The most likely explanation, for this reason, may be that eukaryotic genes often incorporate introns while prokaryotic genes do not possess such a structural arrangement.
<h3>What do you mean by Gene?</h3>
A gene may be defined as a stretch of DNA that contains genetic information that assists in the production of functional protein.
The type of protein may change when there will be a change in the codons. A eukaryotic gene contains introns that are removed during splicing and the codon that codes for specific amino acid may form.
But in prokaryotic genes, no introns are there, and no splicing mechanism will occur, which leads to the formation of different codons, and finally, an alteration in protein may clearly be observed.
Therefore, the most likely explanation, for this reason, may be that eukaryotic genes often incorporate introns while prokaryotic genes do not possess such a structural arrangement.
To learn more about Introns, refer to the link:
brainly.com/question/26464408
#SPJ4
Answer:
B.) Gap junction
Explanation:
Plasmodesmata (singular form: plasmodesma) are intercellular organelles found only in plant and algal cells. (The animal cell "equivalent" is called the gap junction.)
Answer:
B
Explanation:
We never see short bristle males, suggesting some type of lethality. I.e. any males who inherit the mutation die before birth so we don't see the phenotype. This also hints that it could be X-linked.
Females can be short bristled, but males can't, as it is likely lethal. This suggests that having one copy of the short bristle trait without the long bristle trait is lethal (as males as XY and so only have one copy of the trait). The female then must be heterozygous for the short bristle trait (which also explains how in generation F2, long bristle males can be produced, as if she was homozygous males would all be short bristled, and therefore dead, so there would be no males.
Since the first short bristle female is heterozygous, the trait for short bristles must be dominant.
However, since evidence suggests the trait is X-linked, it cannot be autosomal, as suggested in B.
E. <span> When they stop dividing, they do so at random points in the cell cycle; they are not subject to cell cycle controls; and they do not exhibit density-dependent inhibition when growing in culture.
Hope this helps(:</span>
Answer:
100% black
Explanation:
<em>The allele for black is dominant over that of the grey fur.</em>
Let B represents the allele for black color and b for grey color.
Homozozygous black rabbit will have the genotype: BB
Heterozygous black rabbit will have the genotype: Bb
If the two are crossed;
BB x Bb
Offspring: BB, Bb, BB, and Bb = 2BB and 2Bb
2BB = Black fur
2Bb = Black fur (heterozygous)
Both BB and Bb offspring are black phenotypically.
<em>Hence, all the offspring will appear black, a phenotypical ratio of </em><em>100% black.</em>
