Nokea
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.
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Which of the following is most likely to develop sickle cell anemia?
- A. European American
- B. Native American
- C. African American
- D. Asian American
Correct Answer: C
Rationale: Sickle cell anemia is most commonly found in individuals of African American descent. This is because sickle cell trait provides some protection against malaria, and historically, regions where malaria is or was prevalent have higher rates of sickle cell anemia. Therefore, individuals with African ancestry are at a higher risk of developing sickle cell anemia compared to other populations. Choices A, B, and D are less likely to develop sickle cell anemia due to lower genetic prevalence in their respective populations.
A newborn with a respiratory rate of 40 breaths per minute at one minute after birth is demonstrating cyanosis of the hands and feet. What action should a nurse take?
- A. Assess bowel sounds.
- B. Continue to monitor.
- C. Assist with intubation.
- D. Rub the infant's back.
Correct Answer: B
Rationale: Cyanosis of the hands and feet, known as acrocyanosis, is common in newborns shortly after birth and usually resolves on its own. It is not indicative of a need for immediate intervention. Therefore, the appropriate action is to continue monitoring the newborn's condition. Assessing bowel sounds (Choice A) is not relevant to the presenting issue of cyanosis and respiratory rate. Assisting with intubation (Choice C) is an invasive procedure that is not warranted based on the information provided. Rubbing the infant's back (Choice D) is not necessary for acrocyanosis and could potentially disturb the newborn.
A client comes to the clinic for her first prenatal visit and reports that July 10 was the first day of her last menstrual period. Using Nagele's Rule, the nurse calculates the estimated date of birth for the client to be _________.
- A. 4/17.
- B. 4/10.
- C. 5/10.
- D. 5/17.
Correct Answer: A
Rationale: Nagele's Rule is a common method used to estimate the due date. To calculate it, subtract 3 months and add 7 days to the first day of the last menstrual period. In this case, if the last menstrual period started on July 10, subtracting 3 months (April) and adding 7 days gives an estimated due date of April 17. This is the correct answer. Choices B, C, and D are incorrect because they do not follow the Nagele's Rule calculation method.
Females with Turner syndrome:
- A. possess more thymine than cytosine.
- B. are taller than average.
- C. produce little estrogen.
- D. are more likely to give birth to twins.
Correct Answer: C
Rationale: Turner syndrome is a chromosomal disorder in females characterized by short stature and underdeveloped ovaries, resulting in low estrogen production. This leads to symptoms such as delayed puberty and infertility. Choice A is incorrect because the chromosomal abnormality in Turner syndrome does not affect the thymine-cytosine ratio. Choice B is incorrect as females with Turner syndrome are typically shorter than average. Choice D is incorrect as Turner syndrome does not increase the likelihood of giving birth to twins.
When reviewing the electronic medical record of a postpartum client, which of the following factors places the client at risk for infection?
- A. Meconium-stained amniotic fluid
- B. Placenta previa
- C. Midline episiotomy
- D. Gestational hypertension
Correct Answer: C
Rationale: The correct answer is C: Midline episiotomy. An episiotomy is a surgical incision made during childbirth to enlarge the vaginal opening. This procedure increases the risk of infection in the postpartum period due to the incision site being a potential entry point for pathogens. Meconium-stained amniotic fluid (choice A) is a risk factor for fetal distress but does not directly increase the mother's risk of infection. Placenta previa (choice B) is a condition where the placenta partially or completely covers the cervix, leading to potential bleeding issues but not necessarily an increased risk of infection. Gestational hypertension (choice D) is a hypertensive disorder that affects some pregnant women but is not directly associated with an increased risk of infection in the postpartum period.