A newborn baby has become the youngest coronavirus death in the United States after his mother contracted the virus and entered premature labor.
The baby girl, who died on Monday, was born prematurely in Louisiana after her mother was admitted to hospital for symptoms of coronavirus last week.
Although the infant has not tested positive for the virus, authorities anticipate death from the coronavirus due to “the positive nature of the mother,” Dr William “Beau” Clark, parish coroner, said on Monday. ‘East Baton Rouge, during a teleconference. .
“Unfortunately, her pregnancy, which she was carrying, she entered preterm labor and finally gave birth prematurely, and in doing so, the baby, due to the extreme prematurity, did not survive,” he said. -he declares.
“The child has not yet tested positive for COVID-19, however, the mother was.
“In talking to the state epidemiologist, we all agree that I would be included and the doctors involved in the care, that it would be a death related to COVID-19 due to the positive nature of the mother. “
The mother, connected to a ventilator, suffered from shortness of breath and lack of oxygen associated with the virus.
If she hadn’t suffered from these symptoms, she wouldn’t have given birth prematurely, according to Dr. Clark.
As of Monday, at least 477 people died of the virus nationwide, with at least 13,010 confirmed cases – the fourth largest number of confirmed cases in the United States.
More than 9,000 people have died from the virus across the country.
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Although infant coronavirus deaths are rare, the little girl in Louisiana is not the first to succumb to the deadly virus.
A six-week-old baby who tested positive for coronavirus died in Connecticut last Wednesday, Governor Ned Lamont said.
The child’s death is probably one of the youngest of the disease “anywhere,” said Lamont.
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