Nursing Elites

1. After mitosis, the genetic code is identical in new cells unless _________ occur through radiation or other environmental influences.

• A. reductions
• B. expulsions
• C. conceptions
• D. mutations

Correct answer: D

Rationale: After mitosis, the genetic code is typically preserved and remains identical in the new cells. However, mutations can occur due to radiation or environmental influences, leading to changes in the DNA sequence and potentially altering the genetic code. Therefore, the correct answer is 'mutations.' Choices A, B, and C are incorrect because reductions, expulsions, and conceptions do not accurately describe the changes in the genetic code that can result from external factors. Mutations are the only option that reflects the alteration in the genetic code caused by external influences, making it the correct choice in this context.

2. What nursing diagnosis is the most appropriate for a woman experiencing severe preeclampsia?

• A. Risk for injury to mother and fetus, related to central nervous system (CNS) irritability.
• B. Risk for altered gas exchange.
• C. Risk for deficient fluid volume, related to increased sodium retention secondary to the administration of magnesium sulfate.
• D. Risk for increased cardiac output, related to the use of antihypertensive drugs.

Correct answer: A

Rationale: The most appropriate nursing diagnosis for a woman experiencing severe preeclampsia is 'Risk for injury to mother and fetus, related to central nervous system (CNS) irritability.' Severe preeclampsia poses a significant risk of injury to both the mother and the fetus due to complications such as seizures, stroke, and placental abruption. 'Risk for altered gas exchange' is not the priority diagnosis as pulmonary edema is more common in severe preeclampsia. 'Risk for deficient fluid volume' is incorrect as sodium retention in severe preeclampsia often leads to fluid overload. 'Risk for increased cardiac output' is also incorrect as antihypertensive drugs are used to reduce cardiac output in this condition.

3. Which of the following statements is true of a sperm cell?

• A. A sperm cell is significantly larger than an egg cell.
• B. A sperm cell contains two Y chromosomes.
• C. A sperm cell travels randomly inside a woman’s reproductive tract.
• D. A sperm cell is one of the smallest types of cells in the body.

Correct answer: D

Rationale: The correct answer is D. A sperm cell is indeed one of the smallest cells in the human body, much smaller than an egg cell. Choice A is incorrect as sperm cells are much smaller than egg cells. Choice B is incorrect because a sperm cell carries either an X or a Y chromosome, not two Y chromosomes. Choice C is incorrect since sperm cells move in a purposeful manner guided by various factors within a woman's reproductive tract, not randomly.

4. When assessing a woman in the first stage of labor, which clinical finding will alert the nurse that uterine contractions are effective?

• A. Dilation of the cervix.
• B. Descent of the fetus to –2 station.
• C. Rupture of the amniotic membranes.
• D. Increase in bloody show.

Correct answer: A

Rationale: During the first stage of labor, effective uterine contractions lead to cervical dilation. Dilation of the cervix is a key indicator that uterine contractions are progressing labor. Descent of the fetus to -2 station (Choice B) is related to the fetal position in the pelvis and not a direct indicator of uterine contraction effectiveness. Rupture of the amniotic membranes (Choice C) signifies the rupture of the fluid-filled sac surrounding the fetus and does not directly reflect uterine contraction effectiveness. An increase in bloody show (Choice D) can be a sign of impending labor, but it is not a direct indicator of uterine contraction effectiveness.

5. According to a study in the year 2013 by Fellman, if a woman is a twin, if her mother was a twin, or if she has previously borne twins, then:

• A. she will bear only monozygotic (MZ) twins.
• B. the chances of her becoming pregnant decrease.
• C. she is likely to be a healthy mother.
• D. the chances rise that she will bear twins.

Correct answer: D

Rationale: According to the study, the chances of a woman bearing twins increase if she is a twin herself, if her mother was a twin, or if she has previously borne twins. Therefore, the correct answer is D. Choice A is incorrect because the study does not specify that she will bear only monozygotic twins. Choice B is incorrect as the study does not mention any decrease in the chances of becoming pregnant. Choice C is incorrect because the study does not provide information about the woman's health status, focusing instead on the likelihood of bearing twins.

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