Final answer:
Vitamin E is the nutrient that protects polyunsaturated fatty acids from oxidative damage by free radicals. It is a lipid-soluble compound with potent antioxidant properties and is vital for maintaining cellular integrity.
Step-by-step explanation:
Vitamin E protects polyunsaturated fatty acids from being damaged by free radicals. Vitamin E, also known as a-tocopherol, is a lipid-soluble compound that possesses antioxidant properties. As an antioxidant, vitamin E stabilizes free radicals, thus preventing oxidative damage to important cellular components such as the lipid-rich cell membranes. Polyunsaturated fatty acids (PUFAs), which are a type of unsaturated fats found in high concentrations in cell membranes, are particularly susceptible to damage by free radicals. The presence of vitamin E helps to maintain the integrity of these fatty acids, ensuring that they continue to function properly in the body.
Furthermore, the requirement of vitamin E depends on the consumption of PUFAs in the diet and also on selenium status. Foods rich in vitamin E include nuts, seeds, and vegetable oils. A deficiency in vitamin E can lead to various health issues, including muscle weakness and neurological problems, whereas excessive intake, although quite rare, may cause nausea and other symptoms.