a woman gave birth to a 7 pound 6 ounce infant girl 1 hour ago the birth was vaginal and the estimated blood loss ebl was 1500 ml when evaluating the

HESI LPN

HESI Maternal Newborn

1. A woman gave birth to a 7-pound, 6-ounce infant girl 1 hour ago. The birth was vaginal and the estimated blood loss (EBL) was 1500 ml. When evaluating the woman’s vital signs, which finding would be of greatest concern to the nurse?

A. Temperature 37.9°C, heart rate 120 beats per minute (bpm), respirations 20 breaths per minute, and blood pressure 90/50 mm Hg.
B. Temperature 37.4°C, heart rate 88 bpm, respirations 36 breaths per minute, and blood pressure 126/68 mm Hg.
C. Temperature 38°C, heart rate 80 bpm, respirations 16 breaths per minute, and blood pressure 110/80 mm Hg.
D. Temperature 36.8°C, heart rate 60 bpm, respirations 18 breaths per minute, and blood pressure 140/90 mm Hg.

Correct answer: A

Rationale: An estimated blood loss (EBL) of 1500 ml following a vaginal birth is significant and can lead to hypovolemia. The vital signs provided in option A (Temperature 37.9°C, heart rate 120 bpm, respirations 20 breaths per minute, and blood pressure 90/50 mm Hg) indicate tachycardia and hypotension, which are concerning signs of hypovolemia due to excessive blood loss. Tachycardia is the body's compensatory mechanism to maintain cardiac output in response to decreased blood volume, and hypotension indicates inadequate perfusion. Options B, C, and D do not exhibit the same level of concern for hypovolemia. Option B shows tachypnea, which can be a result of pain or anxiety postpartum. Option C and D have vital signs within normal limits, which are not indicative of the body's response to significant blood loss.

2. Which of the following statements about Rh incompatibility is true?

A. Rh incompatibility occurs most commonly during a woman’s first pregnancy.
B. Rh incompatibility is an untreatable condition that leaves a woman infertile for the rest of her life.
C. Rh incompatibility is an abnormality that is transmitted from generation to generation and carried by a sex chromosome.
D. Rh incompatibility occurs due to antibodies transmitted to a fetus during subsequent deliveries causing brain damage or death.

Correct answer: D

Rationale: Rh incompatibility occurs when the mother's antibodies attack the fetus's red blood cells, leading to serious complications, usually in subsequent pregnancies. Choice A is incorrect because Rh incompatibility often occurs in subsequent pregnancies, not necessarily the first one. Choice B is incorrect as Rh incompatibility does not render a woman infertile but can lead to complications during pregnancies. Choice C is incorrect as Rh incompatibility is not carried by a sex chromosome but involves the Rh factor on red blood cells.

3. Before meiosis, a sperm cell:

A. contains 46 chromosomes.
B. contains two X chromosomes.
C. is significantly larger than an egg cell.
D. contains both an X and a Y chromosome.

Correct answer: A

Rationale: Before meiosis, a sperm cell contains 46 chromosomes. This is because sperm cells, like other somatic cells, have a diploid number of chromosomes. During meiosis, the number of chromosomes is halved to 23 to combine with an egg cell during fertilization. Choice B is incorrect because a sperm cell carries either an X or a Y chromosome, not both (Choice D). Choice C is incorrect as sperm cells are generally smaller than egg cells, which is an adaptation that aids in motility and penetration of the egg during fertilization.

4. 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.

5. Jill bears the genetic code for Von Willebrand disease, but she has never developed the illness herself. Jill would be considered:

A. a carrier of the recessive gene that causes the disease.
B. susceptible to the disease after adolescence.
C. an acceptor of the recessive gene that causes the disease.
D. susceptible to the disease in late adulthood.

Correct answer: A

Rationale: Jill is a carrier of the recessive gene for Von Willebrand disease. Being a carrier means that she has one copy of the gene but does not show symptoms of the disease. Carriers can pass on the gene to their offspring. Choice B is incorrect as being a carrier does not mean she is susceptible to developing the disease after adolescence. Choice C is incorrect as 'acceptor' is not a term used in genetics in this context. Choice D is incorrect as susceptibility to the disease is not related to late adulthood in carriers of a recessive gene.