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Table 5 Distribution of fungal isolates with respect to currency type, denomination and source

From: Public health implications of contamination of Franc CFA (XAF) circulating in Buea (Cameroon) with drug resistant pathogens

Currency denomination Fungi isolated N (%)
Absidia sp Acremonium sp Aspergillus sp Blastomyces sp Epidermophyton sp Fusrium sp Scopulariopsis sp Sporothrix sp Microsporum sp Candida albicans Candida sp Penicillium sp Rhyzopus sp Saccharomyces cerevisiae Trichophyton sp
Note 500 0 (0.0) 0 (0.0) 5 (2.3) 0 (0.0) 0 (0.0) 0 (0.0) 0(0.0) 0 (0.0) 0 (0.0) 0 (0.0) 2 (0.9) 4 (1.8) 3 1.4) 1(0.5) 2 (0.9)
1000 0 (0.0) 0 (0.0) 11 (5.1) 2 (0.9) 2 (0.9) 0 (0.0) 0 (0.0) 2(0.9) 2 (0.9) 2 (0.9) 4 (1.8) 11 (5.1) 5 (2.3) 8 (3.7) 6 (2.8)
2000 1 (0.5) 0 (0.0) 10 (4.7) 0 (0.0) 0 (0.0) 0 (0.0) 1 (0.5) 0 (0.0) 0 (0.0) 2 (0.9) 6 (2.8) 10 (4.6) 5 (2.3) 4 (1.8) 3 (1.4)
5000 0 (0.0) 0 (0.0) 3 (1.4) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 1 (0.5) 0 (0.0) 1 (0.5) 0 (0.0) 1 (0.5)
10000 0 (0.0) 1 (0.5) 1 (0.5) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 1 (0.5) 1 (0.5) 1 (0.5) 0 (0.0) 3 (1.4) 3 (1.4) 0 (0.0)
Coin 50 0 (0.0) 0 (0.0) 2 (0.9) 0 (0.0) 1 (0.5) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 1 (0.9) 3 (1.4) 0 (0.0) 1 (0.5) 0 (0.0)
100 0 (0.0) 0 (0.0) 2 (0.9) 0 (0.0) 1 (0.5) 1 (0.5) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 4 (1.8) 5 (2.3) 2 (0.9) 6 (2.8) 5 (2.3)
500 0 (0.0) 0 (0.0) 3 (1.4) 0 (0.0) 1 (0.5) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 2 (0.9) 0 (0.0) 1 (0.5) 0 (0.0) 0 (0.0) 1 (0.0)
Total 1 (0.5) 1 (0.5) 37 (17.3) 2 (0.9) 5 (2.3) 1 (0.5) 1 (0.5) 2 (0.9) 3 (1.4) 7 (3.2) 19 (8.9) 34 (15.9) 19 (8.9) 23 (10.7) 18 (8.4)
χ 2 -test χ2= 3.996 df = 7 P = 0.780 χ2= 12.525 df = 7 P = 0.085 χ2= 9.490 df = 7 P = 0.219 χ2= 7.042 df = 7 P = 0.425 χ2= 5.969 df = 7 P = 0.543 χ2= 4.893 df = 7 P = 0.673 χ 2 =3.996 df = 7 P = 0.780 χ 2 =7.042 df = 7 P = 0.425 χ2= 7.576 df = 7 P = 0.371 χ2= 12.617 df = 7 P = 0.082 χ2 = 3.367 df = 7 P = 0.849 χ2= 10.061 df = 7 P = 0.185 χ2= 6.875 df = 7 P = 0.442 χ2= 9.170 df = 7 P = 0.241 χ2= 6.132 df = 7 P = 0.524
Source of currency                
Butchers 1 (0.5) 0 (0.0) 8 (3.7) 0 (0.0) 2 (0.9) 0 (0.0) 0 (0.0) 2 (0.9) 2 (0.9) 0 (0.0) 5 (2.3) 2 (0.9) 5 (2.3) 4 (1.8) 6 (2.8)
Food vendors 0 (0.0) 0 (0.0) 12 (5.6) 2 (0.9) 1 (0.5) 1 (0.5) 1 (0.5) 0 (0.0) 1 (0.5) 3 (1.4) 4 (1.8) 12 (5.6) 5 (2.3) 8 (3.7) 8 (3.7)
Store attendants 0 (0.0) 0 (0.0) 1 (0.5) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 1 (0.5) 4 (1.8) 4 (1.8) 1 (0.5) 1 (0.5)
Students 0 (0.0) 1 (0.5) 7 (3.2) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 1 (0.5) 10 (4.7) 0 (0.0) 2 (0.9) 0 (0.0)
Hospital/Clinic 0 (0.0) 0 (0.0) 5 (2.3) 0 (0.0) 2 (0.9) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 3 (1.4) 5 (2.3) 6 (2.8) 5 (2.3) 8 (3.7) 2 (0.9)
Taxi drivers 0 (0.0) 0 (0.0) 4 (1.8) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 0 (0.0) 1 (0.5) 3 (1.4) 0 (0.0) 0 (0.0) 0 (0.0) 1 (0.5)
Butchers 1 (0.5) 0 (0.0) 8 (3.7) 0 (0.0) 2 (0.9) 0 (0.0) 0 (0.0) 2 (0.9) 2 (0.9) 0 (0.0) 5 (2.3) 2 (0.9) 5 (2.3) 4 (1.8) 6 (2.8)
Food vendors 0 (0.0) 0 (0.0) 12 (5.6) 2 (0.9) 1 (0.5) 1 (0.5) 1 (0.5) 0 (0.0) 1 (0.5) 3 (1.4) 4 (1.8) 12 (5.6) 5 (2.3) 8 (3.7) 8 (3.7)
χ 2 -test χ2= 5.567 df = 5 P = 0.473 χ2= 9.450 df = 5 P = 0.150 χ2= 13.880 df = 5 P = 0.031 χ2= 6.233 df = 5 P = 0.398 χ2= 4.197 df = 5 P = 0.650 χ2= 3.103 df = 5P = 0.796 χ2= 3.103 df = 5 P = 0.796 χ2= 11.183 df = 5 P = 0.083 χ2= 7.181 df = 5 =0.304 χ2= 4.872 df = 5 P = 0.560. χ2= 4.822 df = 5 P = 0.567. χ2= 28.674 df = 5 P < 0.001 χ2= 11.081df = 5 P = 0.086 χ2= 6.542
df = 5
P = 0.365
χ2= 12.751
df = 5
P = 0.04
  1. N = Number isolated.