A client's Pap test reveals cervical cells with dysplasia changes. The nurse understands that information obtained during the client's health history and identified as a major risk factor for cervical cancer is:

drinking a glass of red wine with dinner.

infection with human papilloma virus (HPV).

having a PAP test performed annually by her gynecologist.

being married for 21 years to the same person.

Answer and Explanation

The Correct Answer is B

A.   This statement is false in terms of being a major risk factor for cervical cancer. While excessive alcohol consumption may have various health implications, moderate drinking, such as a glass of red wine with dinner, is not directly linked to an increased risk of cervical cancer.

B.   Infection with certain high-risk strains of HPV is a major risk factor for cervical cancer. HPV is a sexually transmitted virus that can cause changes in cervical cells, leading to dysplasia and potentially progressing to cancer if not monitored and treated.
C.   In fact, regular Pap tests are a preventive measure that can help detect cervical changes early before they develop into cancer. They are not a risk factor but rather a protective factor against the progression of cervical dysplasia to cancer.
D.   While sexual history can influence HPV transmission, simply being married to the same person for an extended period does not inherently increase the risk of cervical cancer.

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Correct Answer is A

Explanation

A. The elevated pH and decreased PaCO2 indicate that the respiratory system is causing alkalosis, typically due to hyperventilation (excessive breathing).
B. Metabolic alkalosis would be indicated by a high pH with a normal or high HCO3. Since the HCO3 level is normal and the PaCO2 is low, this does not fit the picture of metabolic alkalosis.
C. Respiratory acidosis would be characterized by a low pH and an elevated PaCO2 due to inadequate ventilation and accumulation of carbon dioxide. Here, the pH is high, and the PaCO2 is low.
D. Metabolic acidosis would be indicated by a low pH with a low HCO3. In this case, the pH is high, and the HCO3 is normal.

Correct Answer is ["14"]

Explanation

154 lbs ÷ 2.2 lbs/kg = 70 kg

20 units/kg * 70 kg = 1400 units

1400 units ÷ 25,000 units/250 mL = 14 mL

Therefore, the nurse will administer 14 mL of Heparin.

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ABG Value	Normal Range
PH	7.35-7.45
PaCO2	35-45 mm Hg
HCO3-	22-26 mEq/L
Base excess	-2 to +2
PaO2	80-95 mm Hg
SaO2	>95%

Blood Cell	Normal Ranges
Red blood cells (mature, circulating)	Male: 451-6.01 million/mm3 Female: 4.01-5.51 million/mm3
Hemoglobin	Adult (15-64 yr) Male 14-17.3 g/dL Female 11.7-155 g/dL
Hematocrit	42%-52% in males 36% -48% in females
Reticulocytes	0.5% -2.5% of total RBC count
White blood cells (total)	4.5-11.1 10/mm3
Neutrophils	59% Bands 3% Segs-56%
Eosinophils	2.7%
Basophils	0.5%
Lymphocytes	34%
Monocytes	4%


Platelets	150,000-450,000/mm3

A client's Pap test reveals cervical cells with dysplasia changes. The nurse understands that information obtained during the client's health history and identified as a major risk factor for cervical cancer is:

Nursing Test Bank

Naxlex Comprehensive Predictor Exams

Correct Answer is A

Explanation

A patient is ordered 20 unit/kg IV bolus of Heparin. The patient weighs 154 lbs. Heparin is supplied 25,000 units/250 mL. How many mL will the nurse administer? (round to the nearest whole number)

Correct Answer is ["14"]

Explanation

Heparin is supplied in bags that have 25,000 units / 250 mL. What is the concentration of Heparin?

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