Answer:
25% heterozygous tall
Explanation:
If we take the F1 generation as parents and let them self-fertilise, we have 4 crosses.
The first one for homozygous tall, then we have 100% AA.
The second and third one for heterozygous tall and we have 25% AA, 50%Aa and 25%aa for each of them.
The last one would be for dwarf, and we'll have 100%aa.
Adding all of them, we'll have
AA = 100 + 25 + 25 = 150%
Aa = 50+50 = 100%
aa = 100 + 25 + 25 = 150%
as we had 4 crosses, so dividing the total percentages by 4, we'll have,
AA = 37.5%
Aa = 25%
aa = 37.5%
:. The percentage of heterozygous tall would be 25%.
Hope it helps:)
Natural selection is basically a way for nature to choose what works and what doesn't work. For example, if an animal was born with webs to swim, but most of its food source can be gotten on land, then the animal will die off until it evolves to go on land. That is how evolution works with natural selection. In the past long ago, millions of years back, animals in the water needed oxygen to breath, but the water started to lose it. The land had air, and then the fish slowly grew legs, and their gills would evolve to lungs. Eventually they would be able to roam the land. Because there was so much oxygen with all of the shrubs, trees, and plants, that allowed the bodies of animals to grow big.
The therapy can find the damaged cancer cells and prevent it from reproducing, or even taking it off the gene so that none person from the next generations will suffer the same.
Answer:
The correct answer will be option-B
Explanation:
The process of transcription transcribes the sequence of the bases of genes present on DNA to mRNA which acts as a messenger molecule for the protein synthesis.
The process of transcription is mediated by the presence of enzymes which includes the main transcription enzyme called RNA polymerase. RNA polymerase attaches nucleotides to RNA strand.
The transcription process initiates in three stages: initiation, elongation and termination.
The initiation process begins when RNA polymerase binds to the promoter nucleotide sequence which is present at the start site or the upstream on DNA towards 5' end.
Thus, Option-B is the correct answer.
