MSP430F6438 Interrupt Latency?

Other Parts Discussed in Thread: MSP430F6438

Some background:  We have an application that needs to get in & out of interrupts very quickly.  In some initial tests, we saw that it was taking far too long to do this.  We're running a MSP430F6438 at 20 MHz, and it takes pretty much exactly one microsecond to get from an infinite loop into the ISR and toggle an I/O line.  Looking at the assembly, it's 6 instructions (entry to ISR), 4 instructions to flip the I/O pin.  That should take half a microsecond with a 20 MHz clock.  We're seeing it take a full microsecond. Trying to figure out what I have configured incorrectly.  Any thoughts would be greatly appreciated.

Below is the code that I used to test this and the trace from the oscilloscope that was hooked up.  Yellow trace is a 1 KHz square wave, pink trace is the MSP430F6438 output line.

#include <msp430.h> 
/*
 * main.c
 */
int main(void) {
 WDTCTL = WDTPW | WDTHOLD; // Stop watchdog timer

//Enable 20 MHz oscillator on XT2
 int timeout = 5000;
 //Enable 20 MHZ oscillator.
 P7SEL |= BIT2 | BIT3;

//Enable XT2 Clock
 UCSCTL6 &= ~XT2OFF;

//Configure FLL to be 1:1 with XT2, sourced by XT2
 UCSCTL3 = SELREF_5;

//Wait for XT2 fault to clear.
 do
 {
 UCSCTL7 &= ~(XT2OFFG | XT1LFOFFG | XT1HFOFFG | DCOFFG);
 SFRIFG1 &= ~OFIFG;
 timeout--;
 }
 while ((SFRIFG1 & OFIFG) && timeout);

//Source SMCLK, MCLK, ACLK from XT2
 UCSCTL4 = SELM_5 | SELS_5 | SELA_5;

 //Wire P4.0 into CCR
 P4SEL |= BIT0;
 P4DIR &= ~BIT0;
 TB0CCTL0 = CM_3 | CCIS_0 | CAP | SCS | CCIE;
 TB0CTL = TBCLGRP_0 | CNTL_0 | TBSSEL_2 | ID_0 | MC_2 | TBCLR;

//Wire P1.1 to be output
 P1DIR |= BIT1;
 P1SEL &= ~BIT1;
 P1OUT &= ~BIT1;

_enable_interrupts();
 for(;;)
 {
 _nop();
 }
}

#pragma vector = TIMER0_B0_VECTOR
__interrupt void Interrupt_Timinig()
{
 P1OUT = (P4IN & BIT0) << 1;
}