Answer:
Background
During the course of a bacterial infection, the rapid identification of the causative agent(s) is necessary for the determination of effective treatment options. We have developed a method based on a modified broad-range PCR and an oligonucleotide microarray for the simultaneous detection and identification of 12 bacterial pathogens at the species level. The broad-range PCR primer mixture was designed using conserved regions of the bacterial topoisomerase genes gyrB and parE. The primer design allowed the use of a novel DNA amplification method, which produced labeled, single-stranded DNA suitable for microarray hybridization. The probes on the microarray were designed from the alignments of species- or genus-specific variable regions of the gyrB and parE genes flanked by the primers. We included mecA-specific primers and probes in the same assay to indicate the presence of methicillin resistance in the bacterial species. The feasibility of this assay in routine diagnostic testing was evaluated using 146 blood culture positive and 40 blood culture negative samples.
Explanation:
Results
Comparison of our results with those of a conventional culture-based method revealed a sensitivity of 96% (initial sensitivity of 82%) and specificity of 98%. Furthermore, only one cross-reaction was observed upon investigating 102 culture isolates from 70 untargeted bacteria. The total assay time was only three hours, including the time required for the DNA extraction, PCR and microarray steps in sequence.
Jaundice is a condition that causes the skin of a newborn baby to turn yellow. This happens so because babies are born with extra red blood cells. After birth, the extra red blood cells break down and release a substance called bilirubin in the baby's blood. When there is too much bilirubin in the blood, the baby becomes jaundiced. <span>This condition may last for </span>3-12 weeks<span> after birth, but as long as bilirubin levels are monitored and the baby is feeding well, it rarely leads to any serious complications.</span>
Answer:
Resting potential
Explanation:
This is because , the resting potential is negative in side of the cell because it has more negative charge inside than outside due to accumulation of alot of sodium and potassium ions. when an axon is at rest in the cell, the anions produce negative charge, the sodium pumps eject sodium out of the cell and potassium is injected in, and the sodium gates and potassium gates are all closed.
Answer:
c. Would have deviated from the 9:3:3:1 phenotypic ratio
Explanation:
<em>If two genes are linked together on the same chromosome, the phenotype of the F2 generation would have deviated from 9:3:3:1.</em>
Two genes whose loci are close on the same chromosome are said to be linked. Linked genes have higher frequency of recombination than genes that are not linked.
<u>Hence, while genes that are not linked assort independently to produce 9:3:3:1 phenotypic ratio at F2, linked genes do not assort independently and the higher frequency of recombination ensures that they standard phenotypic ratio is deviated from.</u>
The correct option is c.
