Hardy-Weinberg Equilibrium help

Question

I am currently working on understanding Hardy-Weinberg Equilibrium problems. I have come upon two examples of a problem in two different Youtube videos which got me confused:

Example 1: 19% of a population shows the dominant phenotype. What is the dominant allele frequency?

The instructor in the video says that we should address 19% as a combination of both dominant homozygous and heterozygous, so:

p^2 + 2pq = 0.19

Example 2: In a given population of velociraptors that assumes Hardy-Winberg principles, brown skin is homozygous dominant, and gray skin is homozygous recessive. If the percentage of brown-skinned Velociraptors is 64%, what percentage of velociraptors are Heterozygous?

In this case, the instructor says that p^2 = 0.64.

What I don't understand is why in this case p^2 = 0.64 and not 2pq + p^2 = 0.64.

Couldn't the heterozygous velociraptors also have brown skin?

Referred videos:

First: (Minute 17) https://youtube.com/watch?v=NS3BjKWHK1s

Second: (Minute 5:20) https://youtube.com/watch?v=IVGEusDdJGk

Answer 1

You are correct and the instructor in the second video should have used p^2 + 2pq = 0.64. The only reason I can think of for leaving out the 2pq term is if the heterozygous individuals do not have brown-skin. This would only happen if the gene controlling skin color displayed codominance (heterozygotes would have patches of brown and grey skin) or incomplete dominance (heterozygotes would be a solid greyish-brown color). In either case you would not need Hardy-Weinberg Equilibrium because you can directly count the number of heterozygotes.

Answer 2

I think what question 2 is trying to imply is that only homozygotes for brown alleles are brown skinned. But if the heterozygote has a different phenotype from the homozygotes, I don't see how it is correct to call brown "dominant" and grey "recessive".