Answer:
B) All microorganisms do not produce glucose via glycolysis,
there are alternate pathways that produce glucose.
Answer:
The mentioned parental types are c+m- and c-m+. Thus, the recombinants will be c+m+ and c-m-.
Now, the given distance between c and m is 8 map units. Thus, the recombinant frequency is 8% or 0.08.
The total recombinants from 1000 plaques will come out to be 80,
Thus, the recombinants of each type will be 40.
Total parental type will be 920, and therefore, each parental type count will be 460.
Thus, expected c+m- = 460, expected c-m+ = 460, expected c+m+ = 40 and expected c-m- = 40.
The answer is D, because it is a form of camouflage and an adaptation to the environment for self defense
Answer:
In strict mode, the narrow-sense heritability is the proportion of the additive genetic variance that contributes to the total of the phenotypic variance. This value can be associated with the inheritance of the a-thalassemia
.
Explanation:
A-thalassemia is a genetic disorder caused by mutations affecting four different genes that encode alpha-globin, thus affecting the hemoglobin production process and, consequently, oxygen transport. The mode of inheritance of the a-thalassemia may be associated with narrow sense heritability since the phenotype is manifested by gradation, i.e., each allele might contribute in similar mode to this genetic condition.
Since two females are unable to reproduce. Therefore, if only two females are still alive, the species will become extinct.
