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genetic screening on them and gather the data from the table are able to do some below # OF DOGS GENOTYPE 45 FIFI FIF2 10 45POST-LABORATORY QUESTIONS 4 L Look back at your dog population from Pre-Lab question #5. What is the most likely micruevolutilook back at your dog population from pre-lab question#5. What is the most likely microevolutionary mechanism working alone that could cause this scenario?

genetic screening on them and gather the data from the table are able to do some below # OF DOGS GENOTYPE 45 FIFI FIF2 10 45 F2F2 Since you have the counts of all three genotypes, on Hardy-Weinberg equilibrium) to determine p. use the below equation (not dependent (2 X N(A1A1))+ N(A1A2) p= 2N Once you know p (and q), determine if the population is in Hardy-Weinberg equilib- rium. If you have a laptop, bring it to lab with you to start the lab activity 6.
POST-LABORATORY QUESTIONS 4 L Look back at your dog population from Pre-Lab question #5. What is the most likely micruevolutiocary mechanise working alone that could cause this scenario? 2. imagine a combination ef microevolutiocary mechanisms other than mutation/se- lection. How do you think these mechanisms eould act in concert? What variables would need to be cunsidered in a combiced madel of the mechanisms you have choser? Duw a bypotietical graph of the chaage in allele frequency over time. Hnt: You have to set sucne starting pacameters to make your graph. 3. Optional Te lean more about microcvolutionary mechanisrms in the real world, zead The Beak of the Finch by Jorathaa Weiner. It's à popular press (easy to read) beok abeut soe of the most efective research into microevolutionary mecha- nisns ever doee
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Answer #1

F1F1 = 45
F1F2 = 10
F2F2 = 45

Total population szie = 100
Frequency of F1, p = [(2*45)+10]/2*100
= 0.5
Frequency of F1, p2 = 0.25

Frequency of F2, 2 = [(2*45)+10]/2*100
= 0.5
Frequency of F2, q2 = 0.25

Frequency of F1F2 = 2*0.5*0.5 = 0.5

However, the given data does not relate to H-W equilibrium.
The deviation from the H-W equilibrium in the given case could be due to many reasons.
One possible reason could be a heterozygote disadvantage.
In the given population, both the homozygotes are in great abundance than the heterozygote.
This could be due to a disruptive selection which selective two opposite ends of a phenotype and diminishing the intermediate phenotypic values.

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