House Speaker Mike Johnson on Wednesday accused Ukraine Ambassador to the U.S. Oksana Markarova of election interference and demanded that Ukrainian President Volodymyr Zelenskyy fire her.
Mr. Johnson wrote in a letter to Mr. Zelenskyy that Ms. Markarova’s move to organize a trip to an ammunition factory in the battleground state of Pennsylvania with Democratic Gov. Josh Shapiro was “clearly a partisan campaign event” meant to benefit Vice President Kamala Harris, the Democratic presidential nominee.
“This shortsighted and intentionally political move has caused Republicans to lose trust in Ambassador Markarova’s ability to fairly and effectively serve as a diplomat in this country,” Mr. Johnson, Louisiana Republican, wrote. “She should be removed from her post immediately.
His demands come after the House Oversight and Accountability Committee opened an investigation into the Biden administration’s use of taxpayer money to fly Mr. Zelenskyy to the swing state.
Mr. Zelenskyy is in the country for the U.N. General Assembly this week and will meet with President Biden and Ms. Harris. Mr. Johnson told reporters on Wednesday that he would not be meeting with the foreign leader.
The speaker noted that Mr. Zelenskyy had previously stated that Ukraine should avoid being “captured by American domestic politics” and “influencing the choices of the American people” ahead of the Nov. 5 election. He said the tour of the munitions facility ran afoul of that objective.
He added that all foreign nations should avoid involvement in American politics.
“Support for ending Russia’s war against Ukraine continues to be bipartisan, but our relationship is unnecessarily tested and needlessly tarnished when the candidates at the top of the Republican presidential ticket are targeted in the media by officials in your government,” he said in the letter. “These incidents cannot be repeated. Thank you for your prompt attention to this matter, and I trust you will take immediate action.”
• Alex Miller can be reached at amiller@washingtontimes.com.
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