An Arkansas resident has tested positive for the Zika virus, the Centers for Disease Control and Prevention (CDC) said on Tuesday, the same day New York City announced a second confirmed case of the disease.
According to a statement from the Arkansas Department of Health, the infected resident recently traveled out of the country and had “a mild case of Zika.”
The mosquito-borne virus was first detected in 1947 in African monkeys, and has since seen several short outbreaks in Africa, Asia and the Pacific Islands.
The recent outbreak began last year in Brazil, leading to several deaths and hundreds of cases of micro-encephalitis, an inflammation of the brain that affects the growth and development of the fetus.
The virus has since spread to 20 countries in Central and South America and the Caribbean.
The Arkansas Department of Health advised anyone traveling to those areas to take precautions against mosquitoes, and urged pregnant travelers to consider postponing the trip.
“Arkansas has the kind of mosquitoes that carry Zika virus, so mosquitoes here in Arkansas can become infected with the virus if they bite someone who has Zika. For this reason, people traveling to countries with Zika should avoid mosquito bites for 10 days after they return,” the statement said.
The CDC said it was adding the US Virgin Islands and the Dominican Republic to its Zika virus travel alerts.
There have been two confirmed cases of the virus this year in New York City. The only other confirmed case there occurred in 2013, but health officials acknowledged that more have likely gone unreported.