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Alexus [3.1K]
3 years ago
8

What result would be expected if an additional stimulus, equal in intensity to the first, were to be applied to the muscle at th

e 60 millisecond (ms) time point? relationship between tension and time during a single muscle twitch. what result would be expected if an additional stimulus, equal in intensity to the first, were to be applied to the muscle at the 60 millisecond (ms) time point? relationship between tension and time during a single muscle twitch. the tension exerted by the muscle would continue to decrease, but at a significantly slower rate than observed without the second stimulus. tension would increase to the same maximum force measured at the beginning of phase
c. the muscle would quickly return to the fully relaxed state of minimum tension. the muscle would increase in tension to a level greater than that measured at the beginning of phase
c?
Biology
1 answer:
horrorfan [7]3 years ago
8 0
Looking at the graph attached to this question, the correct option is this: THE MUSCLE WILL INCREASE IN TENSION TO A LEVEL GREATER THAN THAT MEASURED AT THE BEGINNING OF PHASE C.
Adding more load to a muscle increases the amount of work that a muscle is doing and increased work done by the muscles also results in increased muscle tension.  <span />
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Which of the following belong to the Cilates?
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Answer:

Paramecium.

Explanation:

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Endocrine glands and their hormones

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3 years ago
Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some
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Answer and Explanation:

<em><u>The number of observed individuals</u></em>:

  • AA 42
  • AG 24
  • GG 21

<u><em>Total number of individuals, N</em></u>= 87 = 42 + 24 + 21

<u><em>Allelic frequencies</em></u>:

  • f(p) = (2 x AA + AG)/ 2 x N

       f (p)= (2 x 42 + 24) /2 x 87

       f (p) = (84 + 24) / 174

       f (p)= 108 / 174

      f (p) = 0.62

  • f (q) = (2 x GG + AG)/2 x N

        f (q) = (2 x 21 + 24 )/2 x 87

        f (q) = (42 + 24)/ 174

        f (q) = 66/174

        f (q) = 0.38

p + q = 1

0.62 + 0.38 = 1

<em><u>The expected genotypic frequency:</u></em>

  • F (AA)= 0.62 ² = 0.3844
  • F (AG) = 2 x A x G = 2 x 0.62 x 0.38 = 0.4712
  • F (GG) = 0.38 ² = 0.1444

AA + AG + GG = 0.3844 + 0.4712 + 0.1444 = 1

<u><em>The number of expected individuals</em></u>:

AA= (0.62)² x 87 = 0.3844 x 87 = 33.44

AG= (0.4712) x 87 = 40.99

GG= (0.38)² x 87 = 12.563

<u><em>Total number of expected individuals</em></u> = 33.44 + 40.99 + 12.563 = 87

<u><em>Chi square</em></u>= sum (O-E)²/E

  • AA= (O-E)² /E

        AA=(42 - 33.44) ² / 33.44

        AA= 2.2

  • AB= (O-E)² /E      

        AB= (24 - 40.99)²/ 40.99

        AB=7.04

  • BB=(O-E)² /E

        BB= (21-12.563)²/12.563

        BB= 5.66

<u><em>Chi square</em></u>= sum ((O-E)²/E) = 2.2 + 7.04 + 5.66 = 14.9

<u><em>Degrees of freedom</em></u> = genotypes - alleles = 3 - 1 = 2

p value less than 0.05

There is enough evidence to reject the nule hypothesis. The genotype frequencies are not in equilibrium.

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Answer:

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Explanation:

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What type of biome supports about 70% of all living organisms
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An, Biotic Biome.  :D
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