Answer: To show how different members of a species are related.
Answer:
Microfilaments, Intermediate filaments and microtubules
Explanation:
Three distinct elements make up the cytoskeleton in eukaryotic cells are:
1. Microfilaments or actin filaments which are composed of actin proteins. The functions of those filaments are: muscle contraction (myosin heads move “walk” on actin filaments), the movement of the cell, intracellular transport, maintaince of the cell shape..
2. Intermediate filaments which can be made of vimentins, keratin, lamin, desmin… Their functions are: the maintenance of cell shape, anchoring organelles, structural components of the nuclear lamina, cell-cell and cell-matrix junctions…
3. Microtubules are filaments polymers of alpha and beta tubulin. Their roles are in intracellular transport (associated with motor protein dyneins and kinesins), formation of the axoneme of cilia and flagella, formation of the mitotic spindle.
Answer:
can lead to natural selection
Explanation:
<h2>
</h2><h2>
Unequal reproductive success<em> _can lead to natural selection_____.</em></h2>
This is because if there is unequal reproductive success then one of the organisms is <em>better suited for reproducing, surviving, and passing down their genes to their offspring</em> rather than the organism who is unfit. This is where natural selection comes into play because natural selection is process where <em>i</em><u><em>ndividuals better suited for their environment produce more and have better survivality. </em></u>
So overtime, the organism which is weaker (in sense of fitness) (here fitness relates to reproduction and survival), will not pass as many genes and won't survive, whereas, individual better suited will, hence in a way "<u>Nature selected"</u><u> </u>the better fit organism.
Answer:
B. 0.42
Explanation:
Since the two alleles are at Hardy-Weinberg equilibrium, the sum of the frequency of two alleles will be one. If the frequency of a dominant allele is "p" and that of the recessive allele is "q", then p+q=1
According to the given information, the frequency of one allele in the population is= 0.7
This means that the frequency of the other allele would be= 1-0.7 = 0.3
Frequency of heterozygote in the population =2pq = 2 x 0.7 x 0.3 = 0.42
