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United States Senate election, 1988

United States Senate elections, 1988
United States
1986 ←
November 8, 1988 → 1990

33 of the 100 seats in the United States Senate
51 seats needed for a majority
  Majority party Minority party
  GeorgeJMitchellPortrait.jpg Bob Dole, PCCWW photo portrait.JPG
Leader George Mitchell Bob Dole
Party Democratic Republican
Leader's seat Maine Kansas
Last election 55 seats 45 seats
Seats before 54 46
Seats won 55 45
Seat change Increase 1 Decrease 1
Popular vote 35,137,786 31,151,251
Percentage 52.1% 46.2%
Swing Increase 2.0% Decrease 1.4%
Seats up 18 15

1988 Senate election map.svg

  Democratic hold
  Democratic gain
  Republican hold
  Republican gain

Majority Leader before election

Robert Byrd
Democratic

Elected Majority Leader

George Mitchell
Democratic


1988 Senate election map.svg

Robert Byrd
Democratic

George Mitchell
Democratic

The United States Senate elections, 1988 was an election for the United States Senate in which, in spite of the Republican victory by George H. W. Bush in the presidential election, the Democrats gained a net of one seat in the Senate. A total of seven seats changed hands, with four incumbents being defeated. The Democratic majority in the Senate increased from 54–46 to 55–45.

Summary of the 1988 United States Senate election results

The Democrats captured four Republican seats, which included an open seat in Virginia and the seats of three incumbents, Chic Hecht of Nevada, Lowell P. Weicker, Jr. of Connecticut, and David K. Karnes of Nebraska. These gains were partially offset by the Republican capture of open seats by Trent Lott in Mississippi and Connie Mack III in Florida, and the defeat of incumbent John Melcher of Montana to Conrad Burns.


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