Answer:
Some forms of chromatin modification can be passed on to future generation of cells
Acetylation of histone tails in chromatin allows access to DNA for transcription
DNA is not transcribed when packaged tightly in a condensed form
methylation of histone tails can promote condensation of the chromatin
Explanation:
chromatin modifications that can be passed on includes epigenetic modifications that are heritable changes made to the chromatin structure that does not involve the DNA sequences. Some epigenetic modifications include DNA methylation and Histone modifications. examples of histone modification include acetylation, methylation, phosphorylation, ubiquintylation etc. All these function either in allowing the DNA become more accessible to transcritional factors or vice versa. for exmple, histone tail acetylation encourages unwounding of nucleosomes allowing transcriptional factors to have access to the DNa while histone tails methylation further tightens the nucleosomes promoting condensation of the chromatin.
Plants take in that carbon dioxide<span> from the air, and take in water (H2O) as well. Photosynthesis produces a chemical reaction that creates glucose, or C6H12O6, out of CO2 and H2O.
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There are 4 atoms in ammonia.
Answer:
(a) Frequency of M = 0.64
Frequency of N = 0.04
Frequency of MN= 0.32
(b) Expected frequencies of M = 0.648
Expected frequencies of MN = 0.304
Expected frequencies of N = 0.048
Explanation:
(a) If random mating takes place in the population, then the expected frequencies are
f(L(M)) = p = 0.8
F(L(N)) = q
q= 1 - p
= 1 - 0.8
= 0.2
Frequency of M = p^2 = ( 0.8)^2 = 0.64
Frequency of N = q^2 = (1-p)^2 = (1 - 0.8)^2 = (0.2)^2 = 0.04
Frequency of MN = 2pq = 2 * 0.8 * 0.2 = 0.32
(b)
F = inbreeding coefficient = 0.05
f(L(M)L(M)) = p^2 + Fpq = (0.8)^2 + 0.05 * 0.8 * 0.2 = 0.648
f(L(M)L(N)) = 2 pq - 2Fpq = 2 * 0.8 * 0.2 - ( 2 * 0.05 * 0.8 * 0.2) = 0.304
f(L(N)L(N)) = q^2 + Fpq = (0.2)^2 + ( 0.05 * 0.8 * 0.2) = 0.048
