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// answer was short so added this text //
It is c sir ggggyyhhhhhhh
Mitosis and meiosis are both cell division process except that the former is only done by somatic cells and the latter is done by gametes. In G1 phase or the growth phase of interphase, there is increase in cellular substance without DNA replication therefore DNA will be regarded as x. By the time when S or synthesis phase of interphase is finished, DNA replication is done therefore all phases that follow the S phase will have a DNA of 2x. In this case, the cell in metaphase of meiosis I has double the amount of DNA than the one in G1 phase because this cell already underwent the S phase. In meiosis, after the first division (meiosis I), there is reduction of genetic material into half (hence called reduction division) producing haploid cells. So a cell in the metaphase of meiosis II will have DNA half of that in metaphase of meiosis I thereby regarded as x.
Carrier proteins pick up specific molecules and take them through the cell membrane against the concentration gradient. Examples of active transport include: uptake of glucose by epithelial cells in the villi of the small intestine. uptake of ions from soil water by root hair cells in plants.
a) Evolutionary fitness can be characterized as a person's commitment to the cutting edge's quality pool,which is chiefly founded on the aggregate or the genotype of the individual.This principally identifies with the conceptive achievement of an organic entity.
It is fundamentally determined by the offspring that can endure and pass on the characteristics/traits.
b) Graph is already uploaded in the below attached document.
c) The population 3 will have the most elevated frequency of heterozygotes,45% of the complete population.
d) Since population 2 is as of now having a low number of heterozygotes, the occasion would totally eliminate them from the population or decreases their number
Explanation:
Frequency of animals in population 1 having both alleles 1 and 3
frequency of allele 1 =0.60
frequency of allele 3= 0.25
Frequency of both allele 1 and 3 would be allele 1 + allele 3 frequency x 100
= 0.60+0.25 X 100= 85%
