Answer:
The final product is four gametes, two of them with 5 chromosomes, and the other two with 3 chromosomes each.
Explanation:
If nondisjunction occurs during meiosis 1, a pair of homologous chromosomes fail to separate, and one of the daughter cells will have the two chromosomes while the other cell will not get any chromosome from the pair.
If meiosis 1 occurs normally, but nondisjunction occurs in meiosis 2, sister chromatids fail to separate.
The usual process of meiosis produces four daughter haploid cells (n) from a diploid germ cell (2n). Each daughter cell is haploid because they have half the number of chromosomes of the original one.
If the diploid number of the original cell is 8 (2n=8), then under normal conditions, each haploid daughter cell should have 4 chromosomes (n = 4).
But in the exposed example, one pair of homologous chromosomes experiences nondisjunction during meiosis I (in the attached file, you will recognize this pair as the red one). The other chromosomes separate as usual. So one of the daughter cells will have one extra chromosome than expected (five instead of four), and the other daughter cell will lack one chromosome (three instead of four). Meiosis II occurs normally. The final result is the formation of four gametes, two of them with 5 chromosomes, and the other two with 3 chromosomes each.
Answer: c. signal amplification
Explanation:
The uncoupling of the G-protein and inhibiting of the signal amplification are the two affects of the pertusis toxin. It is released by the bacteria called <em>Bordetella pertusis. </em>The G-proteins are affected by the action of pertusis toxin. The production to the excess level of cAMP due to the conversion of ATP into cAMP the ribosylation of the ADP molecules occurs due to pertusis toxin. This leads to the damage of G-proteins.
The answer would be , the sum of all genetic traits in a population’s individuals at any given time
Answer:
True
Explanation:
There are certain limitations associated with the Punnett Squares. They are not useful in case of complex genetic inheritance such as linkage between two genes. In case of linkages it becomes difficult to estimate the distribution of genotypes and phenotypes. For example in case of Nail-patella Syndrome and gene associated with blood group two genes lie on the same chromosome in close vicinity and hence there are high chances of inheritance of these traits in the offspring from the parent thereby causing random distribution of the two traits. This random distribution cannot be captured through punnet square.
The same problem is associated in case where a single gene is determined by multiple genes with graded effects of each gene.
oxygen can only hold two bonds with other elements.
