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BartSMP [9]
3 years ago
9

Ribbon-like acoelomates with bilateral symmetry are examples of _____.

Biology
2 answers:
Anika [276]3 years ago
8 0

Answer:

The correct answer is platyhelminthes: flatworms.

Explanation:

Platyhelminthes are generally known as tapeworms or flatworms, these species are ribbon-like and are soft-bodied invertebrates. Some of the species of this phylum live as parasites in animals and human beings. The most distinguishing characteristic of these invertebrates is their flat and ribbon-like appearance.  

Some of the characteristics of Platyhelminthes are that their body is dorsoventrally flattened, they show radial symmetry, they do not possess a body cavity and thus are known as acoelomate, they are triploblastic comprising three germ layers, and their body is soft and unsegmented.  

Marianna [84]3 years ago
4 0
The answer is <span>Platyhelminthes: flat worms
</span>
Among those groups, cnidarians have radial symmetry so this choice is incorrect.
Among the rest groups, only Platyhelminthes are acoelomates, while other groups are coelomates. Thus, Ribbon-like acoelomates with bilateral symmetry are examples of Platyhelminthes.
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Sixty flowering plants are planted in a flowerbed. forty of the plants are red-flowering homozygous dominant. twenty of the plan
myrzilka [38]

Answer:

The flowering population is in Hardy Weinberg equilibrium.

Step by Step Explanation:

Total initial population size = 60

As both the populations are homozygous so the frequency of alleles will be,

P = 80/120= 0.66

q = 40/120= 0.33

The predicted frequencies for genotypes once the population has reached Hardy-Weinberg

p2 = 0.4356

2pq= 0.4356

q2 = 0.1089

The number of plants with each type of flower in a papulation of 420 is,

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heterozygous= 185

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Chi square analysis:

The observed values for red-flowered plants, pink-flowered plants, and white-flowered plants are not significantly different from the expected values predicted by Hardy Weinberg equilibrium.

Phenotype      observed(o)    expected (e)          (o-e)2/e

Red                        178                 185                       0.26

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White                     52                   50                     0.08

                                       chi-square = 0.474

With 2 degree of freedom this chi-square gives a p value of 0.7 - 0.8, which is not significant.  

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