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2 years ago

The primary motor cortex for control of voluntary muscles is found in the

1 answer:

5 0

Answer: The primary motor cortex for control of voluntary muscles is found in the precentral gyrus of the frontal lobes.

Explanations:

The primary motor cortex is one of the important brain areas involved in motor function. It is found in the precentral gyrus of frontal lobes. It control voluntary muscles and generate impulses needed for movement execution.

Voluntary muscles are muscles that we can control consciously or we can control them at will and we can choose when we want to use them. They are also refers to as skeletal muscles and are attached to bones. The are majorly use for locomotion.

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What the cell of the fish?

kari74 [83]

Fish have <span>Eukaryotic cells

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2 years ago

In a genetic cross between two flowering plants that are heterozygous for the color trait, red flowers (R) are dominant to white

Sedaia [141]

Draw a punnet square:
R. r.
R. RR. Rr
r. Rr. rr

Any combo that results in an R results in red flowers.

3/4 offspring have red flowers.

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3 years ago

Cell reproduction is MOST applicable to which of the following characteristics of life?

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3 years ago

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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

irakobra [83]

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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2 years ago

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