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FinnZ [79.3K]
2 years ago
15

Which part of the brain receives messages from the hair-like receptors

Biology
1 answer:
KatRina [158]2 years ago
4 0
Cerebellum

I hope that helped :)
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Is a male with genotype tt a pct taster
jok3333 [9.3K]
Yes :) : :): :):) :)  ): ):
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Which one of the following statements is accurate?
s344n2d4d5 [400]
<span>Answer D is not correct, as gymnosprems are used widely for consumer goods such as lumber, soap, and nail polish. Answer C is not correct event though wheat and corn are two of the main foods supplied to people in the world. Answer B is a very subjective question, because some people argue that selective breeding helps agriculture, while others do not. The correct answer is A. The green revolution was a period of scientific discoveries that lead to an increase in the supply of food worldwide.</span>
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Photosynthesis and cellular
wlad13 [49]

Answer:

cellular respiration and photosynthesis work together by taking in energy such as the sun, or Co2 to convert that into energy and oxygen right back into the atmosphere.

Explanation:

6 0
2 years ago
Cystic fibrosis is most common in individuals of Northern European descent, affecting 1 in 3200 newborns. Assuming that these al
IgorLugansk [536]

Answer:

0.0177

Explanation:

Cystic fibrosis is an autosomal recessive disease, thereby an individual must have both copies of the CFTR mutant alleles to have this disease. The Hardy-Weinberg equilibrium states that p² + 2pq + q² = 1, where p² represents the frequency of the homo-zygous dominant genotype (normal phenotype), q² represents the frequency of the homo-zygous recessive genotype (cystic fibrosis phenotype), and 2pq represents the frequency of the heterozygous genotype (individuals that carry one copy of the CFTR mutant allele). Moreover, under Hardy-Weinberg equilibrium, the sum of the dominant 'p' allele frequency and the recessive 'q' allele frequency is equal to 1. In this case, we can observe that the frequency of the homo-zygous recessive condition for cystic fibrosis (q²) is 1/3200. In consequence, the frequency of the recessive allele for cystic fibrosis can be calculated as follows:

1/3200 = q² (have two CFTR mutant alleles) >>  

q = √ (1/3200) = 1/56.57 >>

- Frequency of the CFTR allele q = 1/56.57 = 0.0177  

- Frequency of the dominant 'normal' allele p = 1 - q = 1 - 0.0177 = 0.9823

4 0
3 years ago
What can you assume about the amount of glucose (sugar) that is produced as you move the light closer? Would it go up or down?
OverLord2011 [107]

Answer:

I think the answer will be "up"

Explanation:

3 0
3 years ago
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