Answer:
The correct answer would be C. 0.60
It can be calculated with the help of Hardy-Weinberg equation i.e
p² + q² + 2pq = 1
Also, p + q = 1
Number of homozygous dominant (YY) = 36
Number of homozygous recessive (yy) = 16
Number of heterozygous dominant (Yy) = 48
Total number of plants = 100
Now, frequency of homozygous dominant (YY) = 36/100 = 0.36
Therefore, p² = 0.36
So, p = √0.36 = 0.60
Hence, frequency of dominant allele (Y) = 0.60