ATI RN
Questions on Infertility Questions
Question 1 of 5
What is the primary purpose of Preimplantation Genetic Testing (PGT) in IVF?
Correct Answer: B
Rationale: The correct answer is B: To identify genetic abnormalities in embryos before implantation. PGT is used to screen embryos for genetic disorders before implantation in IVF to prevent the transmission of genetic diseases to the offspring. By identifying abnormalities, only healthy embryos are selected for transfer, increasing the chances of a successful pregnancy. Other choices are incorrect as PGT does not directly impact the number of embryos available (A), uterine receptivity (C), or egg production (D) during IVF.
Question 2 of 5
The nurse is teaching a patient about Follistim AQ, a medication prescribed for ovulation induction. What information should be included?
Correct Answer: A
Rationale: Step 1: Follistim AQ is a medication used for ovulation induction. Step 2: The medication contains FSH, a hormone that stimulates follicular development. Step 3: Eggs (oocytes) develop within ovarian follicles during the menstrual cycle. Step 4: Follistim AQ stimulates the growth and development of ovarian follicles. Step 5: Therefore, Option A is correct as it accurately explains how Follistim AQ works. Summary of Incorrect Choices: B: While Follistim is a form of FSH, the key point is its role in stimulating follicular development, not just being a manufactured hormone. C: Rapid weight gain is not a typical effect of ovulation induction with Follistim AQ. D: Side effects like breast tenderness and mood changes may occur, but they are not the primary focus of educating a patient about how the medication works.
Question 3 of 5
A patient undergoing ovarian stimulation reports mild bloating and abdominal discomfort. What is the nurse's appropriate response?
Correct Answer: A
Rationale: The correct answer is A because bloating and abdominal discomfort can be early signs of ovarian hyperstimulation syndrome, a serious condition that requires monitoring. Choice B is incorrect as these symptoms can indeed be related to ovarian stimulation. Choice C is incorrect as abdominal discomfort does not guarantee successful ovulation. Choice D is incorrect as these symptoms do not necessarily suggest treatment ineffectiveness. Monitoring and assessing for signs of ovarian hyperstimulation syndrome are crucial in this scenario.
Question 4 of 5
The nurse educator is conducting a class for student nurses on X-linked recessive disorders. Which responses from a student nurse indicate that further education is needed?
Correct Answer: B
Rationale: The correct answer is B because it is incorrect to assume that if a male doesn't have an X-linked disorder, then his children won't either. This statement overlooks the fact that the male could carry the recessive gene and pass it on to his daughters, who could then become carriers. This demonstrates a lack of understanding of how X-linked recessive disorders are inherited. A: This statement is correct as males cannot be carriers of X-linked recessive disorders. C: This statement is correct as females who are carriers have a 50% chance of passing on the recessive gene to their daughters. D: This statement is correct as sons of carrier females have a 50% chance of inheriting the X-linked recessive disorder.
Question 5 of 5
A patient and her husband request to view the results of their infant's karyotype. An aneuploidy was noted. The nurse understands the results to indicate what?
Correct Answer: A
Rationale: Correct Answer: A Rationale: A karyotype is a visual representation of an individual's chromosomes. An aneuploidy indicates an abnormal number of chromosomes, which can result in genetic disorders such as Down syndrome. A karyotype shows the number, size, and shape of chromosomes, so an abnormal number of chromosomes on the karyogram is the correct interpretation. Summary of Other Choices: B: Translocations involve the movement of genetic material between chromosomes, not an abnormal number of chromosomes. C: Chromosomal breaks would not typically result in an aneuploidy but rather structural abnormalities. D: Duplication of chromosomes to total 46 would not lead to aneuploidy but rather a normal chromosome count.