31–40 of 137 posts
Earlier quoted context omitted.
I hardly think a treatment costing $4000/dose to manufacture is inexpensive.
To cure cancer and live? Seems pretty cheap to me.
And the status of a lot of people's health insurance is suddenly up in the air.
Earlier quoted context omitted.
I don't see it as the author complaining about the danger to the companies but merely stating it. It's not only just OK, but also necessary to throw light on this.
Agreed, but it's still the 3rd of 16 paragraphs. I think it's far the most relevant item.
It's hard to cure cancer definitely because cancer cells will always be trying to out-evolve any treatment. Anything that quickly kills 99% of cells, unfortunately, lets them select for survival traits really fast.
However, given that cancer cells do not evolve between individuals (as opposed to bacteria and virii), if we are to find a efficient immunological treatment it's going to stay efficient. If anything, we will evolve for our cancers to respond better to it.
Happy babies! As a point of interest, this type of engineered T-cell treatment has its roots in research at the Weizmann Institute in Tel Aviv. They demonstrated the use of CARs modified T-cells in curing leukemic disease in rats and mice several years ago. Their research was expanded to human trials at Univ. of Penn. where 27 of 29 patients with incurable leukemia and most of whom had a prognosis of death within a f…
Earlier quoted context omitted.
To cure cancer and live? Seems pretty cheap to me.
Some people don't have $4000 so it's not a matter of deciding whether parting with that much money is worthwhile. And the status of a lot of people's health insurance is suddenly up in the air.
> Newly approved cancer drugs cost an average of $10,000 per month, with some therapies topping $30,000 per month, according to ASCO, which discussed the costs of cancer care at a recent meeting. Just a decade ago, the average cost per month of new drugs was about $4,500.
It seems $4,000 is rather a small fee in comparison.
Earlier quoted context omitted.
To cure cancer and live? Seems pretty cheap to me.
Some people don't have $4000 so it's not a matter of deciding whether parting with that much money is worthwhile. And the status of a lot of people's health insurance is suddenly up in the air.
"The ready-made approach could pose a challenge to companies including Juno Therapeutics and Novartis, each of which has spent tens of millions of dollars pioneering treatments that require collecting a patient’s own blood cells, engineering them, and then re-infusing them." Is author serious with this statement? Two kids lives were saved with a potentially inexpensive technique and the author think's this point is r…
I hardly think a treatment costing $4000/dose to manufacture is inexpensive.
Earlier quoted context omitted.
Agreed, but it's still the 3rd of 16 paragraphs. I think it's far the most relevant item.
Think of it as a warning.
>Although the cases drew wide media attention in Britain, some researchers said that because the London team also gave the children standard chemotherapy, they failed to show the cell treatment actually cured the kids. “There is a hint of efficacy but no proof,” says Stephan Grupp, director of cancer immunotherapy at the Children’s Hospital of Philadelphia, who collaborates with Novartis.
> Although the cases drew wide media attention in Britain, some researchers said that because the London team also gave the children standard chemotherapy, they failed to show the cell treatment actually cured the kids. Could someone explain what the process for proving it would be in this case? I presume some kind of animal testing? Seems pretty immoral to give someone a wildcard treatment and not back it up with so…
Standard approach would likely be to choose a population of patients, randomize them between "chemo" and "chemo+immune" treatment groups, and then test for a significant difference in the metric of choice such as response rate or progression-free survival. The graphical analysis of these data is often done using Kaplan-Meier curves ( https://en.wikipedia.org/wiki/Kaplan%E2%80%93Meier_estimator )