a nurse is caring for a newborn who is 6 hr old and has a bedside glucometer reading of 65 mgdl the newborns mother has type 2 diabetes mellitus which
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1. A nurse is caring for a newborn who is 6 hours old and has a bedside glucometer reading of 65 mg/dL. The newborn’s mother has type 2 diabetes mellitus. Which of the following actions should the nurse take?

Correct answer: D

Rationale: A bedside glucometer reading of 65 mg/dL is within the normal range for a newborn. Reassessing the blood glucose level prior to the next feeding ensures ongoing monitoring without unnecessary intervention. Obtaining a blood sample for a serum glucose level (Choice A) is not necessary as the initial reading is normal. Feeding the newborn immediately (Choice B) may not be indicated and could lead to unnecessary interventions. Administering dextrose solution IV (Choice C) is not warranted as the glucose level is within the normal range and does not require immediate correction.

2. In the prenatal record, the nurse should record for the pregnant client who has a 3-year-old child at home, a term birth, a miscarriage at 10 weeks’ gestation, and a set of twins who died within 24 hours:

Correct answer: C

Rationale: The correct answer is C: 'Gravida 4, para 2.' Gravida refers to the total number of pregnancies, including the current one. In this case, the client has been pregnant a total of 4 times, so gravida is 4. Para is the number of pregnancies that have reached viability, which is 2 in this case. The client has had a term birth and a set of twins who died within 24 hours, totaling 2 pregnancies that reached viability. Choices A, B, and D are incorrect because they do not accurately reflect the client's obstetric history based on the information provided.

3. Twins that derive from a single zygote that has split into two are called:

Correct answer: A

Rationale: The correct answer is A: monozygotic (MZ) twins. Monozygotic twins, also known as identical twins, occur when a single zygote splits into two embryos, leading to two genetically identical individuals. Choice B, fraternal twins, are twins that develop from two separate eggs fertilized by two different sperm cells, resulting in non-identical siblings. Choice C, non-identical twins, is not a common term used to describe this type of twinning. Choice D, dizygotic (DZ) twins, refer to twins that develop from two separate eggs fertilized by two different sperm cells, leading to non-identical twins.

4. When reviewing the electronic medical record of a postpartum client, which of the following factors places the client at risk for infection?

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.

5. _________ is self-propulsion.

Correct answer: C

Rationale: The correct answer is 'Motility.' Motility refers to the ability of cells or organisms to move by themselves, often through the use of specialized structures like flagella or cilia. Mitosis (Choice A) and Meiosis (Choice B) are processes related to cell division and genetic recombination, respectively, not self-propulsion. Mutation (Choice D) refers to changes in the DNA sequence and is not related to self-propulsion.

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