When I try to evaluate a report like this, I try to determine the logic behind it, as well as mathematically calculate what it is trying to prove. I cannot for the life of me wrap my head around the comparison in this report of what amount of fluoride the EPA allows in drinking water and the amount of fluoride found in these foods. It isn't a logical comparison to me. First of all, the amount that is added to drinking water varies from location to location because fluoride is a naturally occurring mineral in water that is not the same in all areas of the country. The purpose of fluoridation is to supplement the water in those areas that do not meet a certain criteria that would prevent tooth decay in that population. Other areas may have too much fluoride in the water, and, therefore, it is filtered at the treatment plant to remove some of it so that it does not exceed what the EPA has determined to be a maximum amount. When deciding how much fluoride should be allowed in drinking water, the EPA also takes into account that we consume a considerable amount of it in our food. We don't live on water alone. As far as I can see, there is no relationship between the amount of fluoride the EPA allows in drinking water and the amount contained in these foods.
There isn't much scientific information that firmly gives what would be a toxic dose of fluoride for dogs and cats. However, I did find one article that mentioned it to be 2 mg/kg of body weight. For my purposes here, I am going to use that figure because it seems reasonable given the toxic dose for humans is considered to be about 5 mg/kg of body weight. Here is a link to an article that discusses toxic and lethal doses of fluoride in adults and children based on an amount per kg of body weight.
http://emedicine.medscape.com/article/814774-overviewIf we have a 5 kg dog (11 lbs), based on the 2 mg/kg of body weight, a toxic dose would be 10 mg. If we take the average of the amount of fluoride found in the dog foods of 8.9 mg/kg of food and calculate how much fluoride is in 1 oz. of that food, the result is 0.25 mg/oz. of food. (1 kg = 2.2lb; 2.2 lb = 35.2 oz; 8.9 mg divided by 35.2 oz = 0.25 mg/oz) If that 5 kg dog consumed 8 oz. of the food in one day, he would consume a total of 2.0 mg of fluoride. Since the toxic dose for that dog would be 10 mg, there is no way he is consuming a toxic dose of fluoride in that food. You can work out the numbers for any size dog eating any amount of food just by plugging in the appropriate numbers. The larger the dog, the more food the dog will eat. But, also, the larger the dog, the larger the toxic dose is going to be if it is 2 mg/kg of body weight. If a very small dog eats a very large bag of food all in one day, there could be a problem. Otherwise, I am not seeing the problem with these foods.
I welcome comments because this is just my logical opinion based on the information I have. I would also welcome any comments on the logic of comparing the amount of fluoride the EPA allows in drinking water with the amounts of fluoride in this food. I can't see the connection, but maybe someone else can.
This post is not meant to be argumentative, but rather an explanation of why I do not believe that the amounts of fluoride in these foods is dangerous. If I am going to say that, then I have to at least present my argument to back it up. I'll gladly admit I am wrong if someone else can come up with additional information. I realize that the 2 mg/kg of body weight that I use here as a toxic dose is unsubstantiated except for what I read in one particular article about oral hygiene in cats and dogs and why we don't use fluoride. It gave this number as the toxic dose. But, it made sense when compared to the numbers given for the human toxic doses that were also based on mg/kg of body weight, and it at least gave a means of calculation that seemed reasonable.
