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Oksana_A [137]
3 years ago
7

While distinctly Anthropoids, some Platyrrhines maintain a few of traits that are typically represented only in Prosimians. For

this question, identify the seemingly primitive traits and discuss how their presence in both Platyrrhines and Prosimians may reflect an example of convergent (parallel) evolution. (150-250 wds)
Biology
1 answer:
crimeas [40]3 years ago
5 0

Convergent evolution is the concept of two or more species evolving similar traits despite being completely unrelated to one another, including the absence of a common ancestor.

The Platyrrhines, more commonly known as "New World Monkeys", share many characteristics with the Prosimians. Some of these include:

  • Teeth layout
  • Diet
  • Eyesight
  • Sleeping habits

and more.

The Platyrrhines and the Prosimians are both nocturnal animals. Despite all other members of their family being active during the day, the Playrrhines have evolved nocturnal habits similar to that of the Prosimians. Aside from that, they<u> feed on insects</u> more so than fruit, which is reflected by the layout of their teeth, and have eyesight well adapted to dark conditions. All of these are traits they share with the Prosimians.

Despite being in different ecosystems and regions, and sharing no recent common ancestors, both of these species <u>have evolved strikingly similar traits.</u> This can be used to corroborate the theory of convergent evolution, which explains how <u>two </u><u>species </u><u>can </u><u>evolve</u><u> very similar </u><u>traits </u><u>despite being unrelated.</u>

To learn more visit:

brainly.com/question/3405872?referrer=searchResults

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You are a botanist looking at the plant color in a new breed of Snapdragon. The two parent plants are Blue and Green and when yo
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Answer:

Incomplete dominance is the inheritance pattern where the dominant allele did not mask the recessive allele completely and form a mix of both alleles. Here the inheritance is the incomplete inheritance. The ratio of F2 generation is 1:2:1.

Given:

R1R1 = 42

R2R2 = 39

R1R2 = 86

Total R1 alleles = 2*42+86 = 170

Total R2 alleles = 2*39+86 = 164

Total alleles = 334

Frequency of allele R1 = 170/334 = 0.51

Frequency of allele R2 = 164 / 334 = 0.49

Expected number of each phenotype:

Total population = 167

Blue = R1R1 = 0.51 * 0.51 * 167 = 43.44

Green = R2R2 = 0.49 * 0.49 * 167 = 40.10

Cyan = 2*R1*R2 = 2*0.51*0.49*167 = 83.46

Phenotype     Observed(O)    Expected (E)    O-E      (O-E)2     (O-E)2/E

Blue                 42                   43.44               -1.44       2.0736     0.0477

cyan                86                   83.46                 2.54     6.4516     0.0773

green              39                    40.1                    -1.1       1.2100      0.0302

Total              167                     167                                               0.1552

Chi-square value = 0.155

Degrees of freedom = no. of phenotypes – 1

Df = 3-1 = 2

Critical value = 5.99

Chi-square value of 0.155 is less than the critical value of 5.99. So we accept the null hypothesis.

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