Answer:
C) hematocrit is highest
Explanation:
The viscosity is the property of the fluid substance related to the flow of the resistance. The flow of the resistance increases when the amount of content in the fluid increases.
The flowing blood contains the water and the other components like the cells and proteins therefore it also exhibit the viscosity.
The viscosity in blood happens when the cells and molecules interact with each other therefore measure of these components indicated about the viscosity of the blood.
The hematocrit is the ratio of the proportion of the red blood cells present in the blood to the total volume of the blood.
The viscosity of blood, therefore, increases when the hematocrit is high and decrease when the hematocrit component is low.
Thus, Option-C is the correct answer.
The reaction force exhibited by the leash is 10N
Answer:
Determine the MDC for the principal diagnosis
Explanation:
MS-DRGs are abbreviated as Medicare Severity Diagnosis Related Groups.
DRGs assignment :
- DRGs are th groups of diagnoses determined by the medicare to be related clinically and have similar resource consumption.
- Many different diagnoses exist within one DRG
- One DRG per hospitalization, assigned at discharge.
- Derived from all diagnoses and procedures listed.
- Groups the patients into categories that consume similar resources
- Measures case mix index
Answer:
1/16
Explanation:
- scalloped (Xsd) is an X-linked recessive trait to Xsd+ (wild type)
- ebony (e) is an autosomal mutation recessive to e+ (wild type).
The two genes are independent because they are located on different chromosomes.
<h3>
<u>Parental generation:</u></h3>
True breeding scalloped female wild type for ebony (<em>Xsd Xsd e+e+</em>) mates with a true breeding male mutant only for ebony (<em>Xsd+ Y ee</em>).
The female only produces <em>Xsd e+</em> gametes. The male produces <em>Xsd+ e</em> or <em>Y e</em> gametes.<u> Therefore, the F1 females will have the genotype </u><u><em>Xsd Xsd+ e e+ </em></u><u>and the F1 males will have the genotype </u><u><em>Xsd Y e e+.</em></u>
If you complete a Punnett Square with the gametes the two F1 individuals can produce, you will get all the F2 proportions. The scalloped, ebony females have a<em> Xsd Xsd e e</em> genotype and appear in a 1/16 proportion.