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Alabama boy now holds record for 'world's most premature baby to survive'

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Posted at 3:31 PM, Nov 11, 2021

An Alabama boy delivered at a gestational age of 21 weeks and one day now holds the title of the “world's most premature baby to survive," according to Guinness World Records.

Guinness announced that baby Curtis Zy-Keith has broken the record on Wednesday.

Initially, Guinness says the baby’s mother’s pregnancy seemed to be progressing well and on track to go to full term, but she was rushed to a hospital for emergency surgery on July 4.

Curtis and his twin sister were born on July 5, 2020, four months ahead of their due date on Nov. 11. Sadly, his twin passed away the next day.

As a newborn, Curtis weighed less than a pound. Doctors said his chances of survival were far less than 1%.

To everyone's amazement, Curtis responded well to treatment, and he grew stronger as the weeks went on, though he needed constant care.

The baby was on a ventilator for three months and in the hospital for 275 days before he was finally able to go home on April 6 of this year.

Guinness says Curtis could only go home thanks to a tailored course of medication and special equipment such as bottled oxygen and a feeding tube, but it was nevertheless a major milestone for the boy.

This past July, Curtis was able to celebrate his first birthday with his family.

The previous baby to hold the record for “most premature baby to survive” was named earlier this year, a boy from Wisconsin named Richard Hutchinson. He was born at a gestational age of 21 weeks and two days on June 5, 2020, exactly a month before Curtis was born.

Prior to Richard, Guinness says this record had remained unbroken for 34 years. That baby was born in Ontario at the gestational age of 21 weeks and five days on May 20, 1987.