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Tcecarenko [31]
2 years ago
6

What were the stimuli that caused the sensors to determine your body moved away from

Biology
1 answer:
san4es73 [151]2 years ago
3 0

Answer:

Body temperature

Explanation:

When it get either too hot or too cold, body temperature changes and regulated by negative feedback. The stimulus is when body temperature increases or decreases from 37 degrees Celsius and the sensors sends stimulus to skin and  the hypothalamus in brain referring that your temperature has strayed from its set point.

Hence, the correct answer is "Body temperature".

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What proportion of people who are infected with polio develop neurological problems?
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Each of the four pedigrees that follow represents a human family within which a genetic disease is segregating. Affected individ
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<u> The following four traits are -: </u>

  • <u>Pedigree 1 -</u> A recessive trait (autosomal recessive)  is expressed by pedigree 1.
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  • <u>Pedigree 3</u> - The inheritance of the dominant trait (autosomal dominant) is illustrated by Pedigree 3.
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Explanation:

<u>Explaination of each pedigree chart</u>-

  • Pedigree 1 demonstrates the <u>recessive trait </u>since their children have been affected by two unaffected individuals. If the characteristics were X-linked, in order to have an affected daughter, I-1 would have to be affected. X^A In this, both parents are autosomal recessive trait carriers, so the child will be affected by a 1/4 (aa)
  • <u> Recessive inheritance</u> is defined by <u>Pedigree 2</u>. This is<u> X-related inheritance as autosomal recessive</u> inheritance has already been accounted for in part 1. This inference is confirmed by evidence showing that the father (I-1) is unaffected and that only the sons exhibit the characteristic in generation II, suggesting that the mother must be the carrier. The individual I-2 is a carrier for this X-linked trait. A typical  Xa chromosome is attached to the unaffected father (I-1), so the chance of carrier II-5 is 1/2. Probability of an affected son = 1/2 (probability II-5 is a carrier) x 1/2 (probability II -5 contributes (X^A) x 1/2 (probability of Y from father II-6) = 1/8. An affected daughter's likelihood is 0 because a typical X^A must be contributed by II-6.
  • The inheritance of the<u> dominant trait</u> is demonstrated by <u>Pedigree 3 </u>because affected children still have affected parents (remember that all four diseases are rare). The trait must be <u>autosomal dominant</u> because it is passed down to the son by the affected father. There is a 1/2 risk that the heterozygous mother (II-5) would pass on mutant alleles to a child of either sex for an autosomal dominant feature.
  • <u>Pedigree 4</u> is an <u>X-linked dominant function</u> characterized by the transmission to all of his daughters from the affected father but none of his son. On the mutant X chromosome, the father (I-1) passes on to all his daughters and none of his sons. As seen by his normal phenotype, II-6 therefore does not bear the mutation. An affected child's likelihood is 0.    

In the question the pedigree chart was missing ,hence it is given below.

     

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impossible without the motel

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