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natta225 [31]
3 years ago
8

How can you explain the bimodal distribution, with most soapberry bug individuals either having a long or a short proboscis seve

ral decades after the introduction of the non-native soapberry species. Focus on what selection pressures may have acted on this population and what the fitness of different individuals of varying phenotypes are in this population.
Biology
1 answer:
Setler [38]3 years ago
7 0

Answer:

need the points

Explanation:

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A new nucleotide can only be added to the _____ end of a growing dna strand. dna therefore always grows in the _____ direction.
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A new nucleotide can only be added to the 3' end of a growing dna strand. dna therefore always grows in the 5' to 3' direction.

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1 year ago
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Ansa million. once you're staring at in ordinary terms one technology, it relatively is, all those fruit flies have been produced via 2 mum and dad, then the only plausible parental genotype that provide you a 50/50 ratio is that if the mummy is XwXw and the father is Xry... this gene is carried on the X chromosome, and w represents white eyes and r represents purple eyes. you are able to bypass forward and draw this out in case you like, even yet it provides 50% heterozygous women individuals (with purple eyes, provided that purple is dominant), and 50% white eyed men, who can in ordinary terms inherit Xw from their mom. The Y that they could inherit from the father could have not any effect on their eye shade. it relatively is comparable for women individuals, they are in a position to in ordinary terms inherit Xw from their mom and an Xr from their father, meaning they'll continually be heterozygous for the attention shade gene. 2. returned, set us the genotypes so which you will visualize it; we are going to say that this disease is carried on the X chrmsm (thats whats usually assumed), and the gene for customary creative and prescient is XB, yet for colorblindness its Xb. those are dominant/recessive to boot. So, if the girl is a provider, her genotype is XBXb, and if the male is colorblind, his genotype is XbY. in case you draw this out in a punett sq., you will get 25% heterozygous woman, 25% colorblind woman, 25% well-known male, and 25% colorblind male. so, there's a 50% risk that any baby they have would be colorblind, 50% risk if its a woman, or 50% risk if it relatively is male.wer:

Explanation:

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