Answer:
The nervous system helps all the parts of the body to communicate with each other. It also reacts to changes both outside and inside the body. The nervous system uses both electrical and chemical means to send and receive messages.
Explanation:
The correct answer is to accurately measure core temperature.
The tube seen entering the reporter's nose is a thermistor which is used to measure the core temperature of the body with very high levels of accuracy. The tube reaches the esophagus, which is one of the most preferred locations for measuring the core temperature.
The core body temperature is the operating temperature of the human body and it refers to the temperature of the deeper structures of the body instead of the peripheral tissues.
The esophagus is close to the left ventricle of the heart, it has a deep body location and a quick response to any changes in the body temperature.
Answer:
the answer for this question is d
IT IS BISEXUAL SINCE IT HAS BOTH MALE AND FEMALE PARTS
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
