Bacterial infection in cows reduced with vitamin D – Oct 2011

Treatment of an intramammary bacterial infection with 25-hydroxyvitamin D(3).

PLoS One. 2011;6(10):e25479. Epub 2011 Oct 3.
Lippolis JD, Reinhardt TA, Sacco RA, Nonnecke BJ, Nelson CD.
Ruminant Diseases and Immunology Research Unit, National Animal Disease Center, Agricultural Research Service, United States Department of Agriculture, Ames, Iowa, United States of America.

Deficiency of serum levels of 25-hydroxyvitamin D(3) has been correlated with increased risk of infectious diseases such as tuberculosis and influenza. A plausible reason for this association is that expression of genes encoding important antimicrobial proteins depends on concentrations of 1,25-dihydroxyvitamin D(3) produced by activated immune cells at sites of infection, and that synthesis of 1,25-dihydroxyvitamin D(3) is dependent on the availability of 25-hydroxyvitamin D(3). Thus, increasing the availability of 25(OH)D(3) for immune cell synthesis of 1,25-dihydroxyvitamin D(3) at sites of infection has been hypothesized to aid in clearance of the infection. This report details the treatment of an acute intramammary infection with infusion of 25-hydroxyvitamin D(3) to the site of infection.

Ten lactating cows were infected with in one quarter of their mammary glands. Half of the animals were treated intramammary with 25-hydroxyvitamin D(3).

The 25-hydroxyvitamin D(3) treated animal showed significantly lower bacterial counts in milk and showed reduced symptomatic affects of the mastitis.

It is significant that treatment with 25-hydroxyvitamin D(3) reduced the severity of an acute bacterial infection. This finding suggested a significant non-antibiotic complimentary role for 25-hydroxyvitamin D(3) in the treatment of infections in compartments naturally low in 25-hydroxyvitamin D(3) such as the mammary gland and by extension, possibly upper respiratory tract infections.

PMID: 21991312

40,000 IU of vitamin D twice daily

cows weigh 600 kg, 200 lb person = 90 kg, so this dose would be 40,000 *(90/600) = 6,000 IU for a 200 lb person
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