Chemistry, 31.05.2021 15:10 bayleeharris8p78txa

A 45-year-old woman comes to a clinic with complaints of morning stiffness in her ankle joints, worse on rising in the morning and improving during the day. Her discomfort is responsive to aspirin. She has also been fatigued and weak. During the last week, she has noticed that her wrist and ankle joints on both sides of her body are also painful and swollen. Blood is drawn to test for rheumatoid factor and antinuclear antibody. Synovial fluid is aspirated and analyzed. Results from analysis of the synovial fluid rule out crystal deposition diseases, such as gout and pseudogout, and no infectious microorganisms are seen. Required:
a. What can be the possible cause of a false-positive RF assay?
b. Do the patient's clinical symptoms and laboratory results support a diagnosis 5. If so, what diagnosis would be the first choice, and why?

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In all the patients with rheumatoid arthritis RF is not present. It can also be witnessed in patients with conditions like hepatitis, bacterial endocarditis, and tuberculosis. Some of the reasons that can also produce false-positive results for RF assay are lipemic, hemolyzed, or heavily contaminated serum, other rheumatic diseases like systemic lupus erythematous, reading test results post the specified time of two minutes, and due to some of the chronic infectious diseases like hepatitis, tuberculosis, and syphilis.

b. The symptoms of the oatient like inflammation and pain in joints and the results of the lab indicate that the synovial fluid is drying and formation of crystal is taking place. These are the conditions that suggests the existence of rheumatoid arthritis.

c. In the given case, the first choice diagnosis would be the blood test that would show the presence of anti-CCP antibodies and the drying of synovial fluid. The presence of anti-CCP antibodies results in wear and tear in the joints, thus showing the diseased conditions.

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

Answer from: Quest

It's called the falling actionwrong subject by the way

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