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A woman is 6 weeks pregnant. she has had a previous spontaneous abortion at 14 weeks of gestation and a pregnancy that ended at 38 weeks with the birth of a stillborn girl. what is her gravidity and parity using the gtpal system?

User Mawardy
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2 Answers

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Final answer:

Using the GTPAL system, the woman's reproductive history would be represented as 3-1-0-1-0: 3 pregnancies in total, 1 term birth that resulted in a stillborn, 0 preterm births, 1 abortion (miscarriage), and 0 living children.

Step-by-step explanation:

The system used to describe a woman's reproductive history is called the GTPAL system. The acronym stands for Gravidity, Term births, Preterm births, Abortions, and Living children. In this case, a woman who is 6 weeks pregnant, has had a previous spontaneous abortion at 14 weeks (considered a second-trimester miscarriage), and a stillbirth at 38 weeks (a Term birth, even though the child was not born alive), can be described using the GTPAL system as follows:

  • Gravidity (G): The total number of times the woman has been pregnant, including current pregnancy. Hence, this will be 3.
  • Term births (T): The number of pregnancies that ended at or after 37 weeks. She has 1 term birth that unfortunately resulted in a stillborn.
  • Preterm births (P): The number of pregnancies that ended after 20 weeks but before 37 weeks. In her case, this is 0 since she does not have any live preterm births.
  • Abortions (A): The number of pregnancies that ended before 20 weeks. This includes both miscarriages and elective abortions. Therefore, this number is 1.
  • Living children (L): The total number of living children. For this woman, it is 0 as she has no living offspring.

So, the GTPAL for this woman would be 3-1-0-1-0.

User Apolfj
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3 votes
so she didn't get abortion then 
User Jack Trowbridge
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