A fruit fly population has a gene with two alleles, A1 and A2. Tests show that 70% of the gametes produced in the population contain the A1 allele. If the population is in Hardy-Weinberg equilibrium, what proportion of the flies carries both A1 and A2

0.42

Explanation:

Since 70% of the gametes have an A1 allele, this means that the frequency of the A1 allele in the population is 70% or 0.7.

The population is in Hardy-Weinberg equilibrium. For this population, the frequency of another allele for the same locus (A2) would be= 1- frequency of A1 allele = 1-0.7 = 0.3

The proportion of flies carrying both A1 and A2 alleles would be = 2 x frequency of A1 allele x frequency of A2 allele.  

So, frequency of heterozygous flies= 2 x 0.7 x 0.3 = 0.42 or 42%

1. a host loses nutrients to a parasite

2. a parasite only benefits when the host is alive (cause if the host dies the parasite dies with it )

3. producers to consumers

4. the tapeworm is a parasite

5. they convert energy from the sun (photosynthesis)

6. sunflower-robin (i think im 80% sure)

7. producer

8. consumer

9. prey

10. fresh water snails ( cause the green algea is the producer and the snails are the primary consumers and the small fish consumer so is the big fish)

idk why hjggjhjkhj