For this question, we will utilize a population of martians that is in hardy weinberg equilibrium. the dominant martian phenotype is the possession of 2 antennae. in this population, 84 of the martians have 2 antennae, while 16 lack antennae. what is the frequency of heterozygotes in this population?
a. o.16
b.0.4
c.0.6
d.0.24
d.0.48

d.0.48

Explanation:

When a population is in Hardy Weinberg equilibrium the genotypic frequencies are:

freq (AA) = p²

freq (Aa) = 2pq

freq (aa) = q²

p is the frequency of the dominant A allele and q is the frequency of the recessive a allele.

In this population of 100 individuals, 84 martians have the dominant phenotype and 16 have the recessive phenotype.

Therefore:

q²=16/100

q² = 0.16

q=√0.16

q = 0.4

And p+q=1, so:

p = 1 - q

p = 1-0.4

p = 0.6

The frequency of heterozygotes is:

freq (Aa) = 2pq = 2 × 0.4 × 0.6

freq (Aa) = 0.48

the cladogram shown here shows similarities and differences among several classes of vertebrates. anatomically, the four have limb structure as their common ancestor. we might separate the mammals from the other three classes because they a) divergent; do not lay eggs b) identical; are warm-blooded c) varied; a four chambered heart d) homologous; have mammary glands

sheep is a herbivore animal whereas grasshopper belongs to the class chordates but both of them feed on greens.  

if grasshopper and sheep feeding on a pasture land, i do not think both of them will have any impact on each other. however, the number of the species can make difference. since sheep feed on greens, there is a possibility there left no pasture land for the grasshoppers or they may get extinct.