A baby girl in India was born with eight embryos inside her stomach, doctors have claimed.
The girl was born at a private hospital in Ramgarh District, Jharkhand State, India on October 10.
Her parents reportedly took her to hospital with concerns that she had a tumour due to her abnormally swelled stomach.
Doctors transferred the girl to the Rani Children Hospital & Research Centre in the city of Ranchi following a CT scan which suggested she had a tumour.
However, it has since been confirmed that the baby had fetus-in-fetu (FIF). It is a rare condition where one or more malformed vertebrate foetuses are enclosed within the body of their sibling.
A total of eight undeveloped embryos were found in the girl's stomach. It's believed to be the highest number ever found in a newborn.
Dr Mohammed Imran confirmed the operation was successfully completed in around one and a half hours.
The baby girl, who has not been identified, is currently recovering and said to be in a good condition.
Such cases are considered extremely rare and they only happen in fewer than one in every 1 million births. Only 200 cases of the extraordinarily rare defect have ever been recorded, Daily Mail reports.
It's been reported in local media that previous cases have only seen two or three embryos removed from the newborn, and never as many as eight.
Foetus-in-fetu is caused by the incomplete separation of twins, which fails to grow and instead becomes an internal part of the healthy twin.
