Chromosome duplications and deletions frequently result in abnormal phenotypes or inviable gametes. Gene dosage is modified is a major contributor to this phenomenon.
In the field of genetics, we can define gene dosage as the quantitative measure or copies of a particular gene that is present in an organism. Abnormalities in the gene dosage at a particular location can cause severe damage to the resulting phenotype.
Gene dosage can lead to chromosome duplications if the copy number or gene product is more and it can cause deletions if the copy number or gene product is less. Such complications will result in abnormal phenotypes or inviable gametes. For example, in Down's syndrome, the person has a modification of the 21st chromosome as there is one extra 21st chromosome present. This leads to a variety of diseases and defects in the person.
Although a part of your question is missing, you might be referring to this question:
Chromosome duplications and deletions frequently result in abnormal phenotypes or inviable gametes. Which factor is a major contributor to this phenomenon?
a. Recessive diseases are unmasked by additional copies.
b. The genes are found in a novel arrangement.
c. Gene dosage is modified.
To learn more about deletions, click here: brainly.com/question/2033407
#SPJ4
A desert?? I think
Hope I helped though
NADPH builds up in the STROMA of the chloroplasts during electron transport. In chloroplasts, NADP+ is reduced by the ferredoxin-NADP+ reductase.
Nicotinamide adenine dinucleotide phosphate (NADPH) is the reduced electron form of NADP+.
The NADPH coenzyme is generated during the last step of the electron transport chain of the light-dependent reactions of photosynthesis.
NADPH is used by plant cells as an electron donor in different reactions associated with dehydrogenase and reductase enzymes.
Learn more in:
brainly.com/question/6001645?referrer=searchResults
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
