Additional exercise (ANOVA)

The first researcher has used a two-way ANOVA, followed by a TukeyHSD, and obtained the following results:

> summary(data.aov) 
                   Df Sum Sq Mean Sq F value Pr(>F) 
genotype           1    2.7    2.72   0.055  0.818 
treatment          1   35.9   35.87   0.720  0.409
genotype:treatment 1   16.9   16.93   0.340  0.568
Residuals         16  797.2   49.83

> TukeyHSD(data.aov) 
  Tukey multiple comparisons of means 
    95% family-wise confidence level 
Fit: aov(formula = measure ~ genotype * treatment, data = data) 

$genotype 
           diff     lwr      upr     p adj 
KO-WT0.7378864 -5.95428 7.430053 0.8181487 

$treatment
        diff       lwr      upr     p adj 
T-C 2.678292 -4.013875 9.370458 0.4087217 

$`genotype:treatment` 
                diff        lwr      upr     p adj 
KO:C-WT:C  2.5779733 -10.194827 15.35077 0.9374098 
WT:T-WT:C  4.5183786 -8.254422  17.29118 0.7448413 
KO:T-WT:C  3.4161779 -9.356623  16.18898 0.8688435 
WT:T-KO:C  1.9404052 -10.832395 14.71321 0.9716065 
KO:T-KO:C  0.8382046 -11.934596 13.61101 0.9975520
KO:T-WT:T -1.1022006 -13.875001 11.67060 0.9944969 

The other researcher was only interested in the difference (within the controls), between the WT and KO. He decided to use a t-test comparing these groups:

> t.test(data$measure[data$treatment=="C"] ~ data$genotype[data$treatment=="C"])

   Welch Two Sample t-test 

data:  data$measure[data$treatment == "C"] by
       data$genotype[data$treatment == "C"]
t = -5.2689, df = 7.487, p-value = 0.0009369
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -3.719856 -1.436091
sample estimates:
mean in group WT mean in group KO 
        1.986127         4.564100 

The results are obviously not coherent: the t-test indicates that the 2 groupes are very significantly different, but neither the ANOVA nor the Tukey HSD results indicate an difference between any two groupes. Can you describe what explains the difference ?  (a good starting point would be to redo the two analyses from scratch).