The following graph represents the stages of labor. Which of the following statements is true about the graph C?
Correct Answer: Secondary arrest after progression of labor
Description: Ans. is a, i.e. Secondary arrest after progression of laborNow friends--first you should know how to read this graph.If you see their is a horizontal line shown which corresponds to 3 cms, i.e. in this graph-active phase is beginning at 3 cms.In the graph line marked as 'mean'--means normal graph.In the graph marked as mean, the patient is reaching 3 cms at 8 hrs, i.e. latent phase ends at 8 hrs and active phase begins. Now in active phase minimum dilatation should be 1 cm/hr i.e. from 3 cms to 10 cms it should take maximum 7 hrs, i.e. 8+7=15 hrs. Now if you see the patient reached 10 cms (full dilatation) well within the time i.e. at 12 hrs.Now keeping this mind let's analyze:Graph AIn graph A--patient is reaching 3 cms i.e. active phase at 20 hrs or more, which means latent phase is prolonged.As prolonged latent phase in:Nulliparous = 20 hrsMultiparous = 14 hrsSo graph A shows prolonged latent phaseGraph BLatent phase (i.e. uptil 3 cms dilatation) occurred at 14 hrs. Then from 3 cms to 10 cms the patient is taking around 10 hours (14 to 24 hrs) whereas this 7 cms of dilation should have been completed in maximum 7 hours 1 cm/hr. This means here active phase is getting prolonged.Graph CThe latent phase ended at 8 hours. Then the patient dilated till 6 cms 10-11 hours.Now from 11 to 16 hours she is 6 cms dilated, i.e. 4 hours have passed and still there is no dilatation. Now since nothing has been mentioned about membranes, I am assuming they are ruptured.So it is a case of active phase arrest i.e. secondary arrest after progression of labor.
Category:
Gynaecology & Obstetrics
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