True, to perform ABG punctures, a phlebotomist must undergo extensive training.
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What is an ABG test?</h3>
- An blood gas (ABG) test measures the oxygen and carbon dioxide levels in your blood as well your blood's pH balance.
- Arterial blood gas tests can help healthcare providers interpret conditions that affect your respiratory system, cardiovascular system and metabolic processes (how your body transforms the food you eat into energy), especially in emergency situations.
- There’s also a test referred to as a "blood gas analysis," which uses a sample of blood from anywhere in your cardiovascular system (artery, vein or capillary).
- An blood gas (ABG) test only tests a blood sample from an artery in your body.
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Answer:
In the most general case of x bases and y bases per codon, the total number of possible codons is equal to xy .
In the case of the hypothetical Martian life-forms, is the minimum codon length needed to specify 17 amino acids is 5 (25 = 32), with some redundancy (meaning that more than one codon could code for the same amino acid). For life on Earth, x = 4 and y = 3; thus the number of codons is 43, or 64. Because there are only 20 amino acids, there is a lot of redundancy in the code (there are several codons for each amino acid).
Explanation:
They depend on the amount of local rain fall.
Answer:
Cellular Fluids and Organelles
Explanation:
Hope this helped
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
