EVPP 110 Lab Succession on a Small Scale (Activity 3B) Microevolution (Activities 1-2) Week of December 3rd 2018 Version 1.1. Last updated: 12/2/2018 2:03:50 PM

Succession on a Small Scale – Activity 3 – Winogradsky Column – Part B – Final Observation

Part B: Final Observation Sketch Figure 3.2. Sketch final appearance of control and treatment Winogradsky columns. Please DO NOT open the columns, regardless of what your textbook tells you! Standard column (control) Treatment column

Microevolution – Introduction

Population All individuals of a given species living in the same geographical area at a given time. Gene pool Genetic make-up of a population. All the genes in a population. Microevolution A change in the gene pool – or in other words, a change in the allele frequencies.

Changes in the environment lead to the selection of genetic traits better fitted to that environment. Better adapted (fit) individuals have greater success in reproduction and pass the favored traits to their offspring. Favored traits accumulate in a population. Populations evolve, not individuals.

When there are no changes in the gene pool, population is in: Hardy-Weinberg equilibrium Requires five conditions: Large population. Random mating among all individuals. (Genetically) isolated population. No genetic mutations occurring. All individuals have equal success in reproduction.

Impossible for all five conditions necessary for Hardy-Weinberg equilibrium to be present. This leads to: Microevolution – a change in the allele frequencies. Five causes: Small population. Non-random mating. Population not isolated. Genetic mutations are occurring. Unequal success in reproduction.

Microevolution – No Natural Selection

Color shade (phenotype) Genotype Phenotype example Dark BB (Homozygous dominant) Medium Bb (Heterozygous) Light bb (Homozygous recessive) Color is assumed to be controlled by a single gene (letter “b”). Inheritance pattern is assumed to be autosomal dominant. Assumed there are two alleles for this gene. Dominant allele represented by “B”, recessive allele represented by “b”.

Sample first mating pair selected: Table 1.1. Total number of mating pairs for each possible genotype combination in a pair Genotype combination of mating pair Tally Marks # Pairs Dark (BB) x Dark (BB) x Dark (BB) x Medium (Bb) or Medium (Bb) x Dark (BB) x or x Dark (BB) x Light (bb) or Light (bb) x Dark (BB) Medium (Bb) x Medium (Bb) Medium (Bb) x Light (bb) or Light (bb) x Medium (Bb) x or x Light (bb) x Light (bb) Total # Pairs: 50 or Create population of 100 moths. Select random pairs, record combinations.

Table 1.1. Total number of mating pairs for each possible genotype combination in a pair Genotype combination of mating pair Tally Marks # Pairs Dark (BB) x Dark (BB) x 1 Dark (BB) x Medium (Bb) or Medium (Bb) x Dark (BB) x or x 9 Dark (BB) x Light (bb) or Light (bb) x Dark (BB) 4 Medium (Bb) x Medium (Bb) 11 Medium (Bb) x Light (bb) or Light (bb) x Medium (Bb) x or x 17 Light (bb) x Light (bb) 8 Total # Pairs: 50 Repeat until you have selected and recorded information for 50 mating pairs. Total the tally marks for each row and record the results. The “# pairs” column should add up to 50 (100 moths grouped as 50 mating pairs).

Assumed # and type of offspring Table 1.2. Number and genotype of offspring of mating pairs Mating pair Assumed # and type of offspring Total # pairs (Table 1.1) # of BB offspring # of Bb offspring # of bb offspring BB x BB 4 BB 1 BB x Bb 2 BB, 2 Bb 9 BB x bb 4 Bb 4 Bb x Bb 1 BB , 2 Bb, 1 bb 11 Bb x bb 2 Bb, 2 bb 17 4 bb 8 Subtotals: Grand total (#BB + #Bb +#bb): Transfer data from “# pairs” column of Table 1.1 to the third column of Table 1.2 (headed “total # of pairs (from Table 1.1)”).

Assumed # and type of offspring Table 1.2. Number and genotype of offspring of mating pairs Mating pair Assumed # and type of offspring Total # pairs (Table 1.1) # of BB offspring # of Bb offspring # of bb offspring BB x BB 4 BB 1 4 BB x Bb 2 BB, 2 Bb 9 18 BB x bb 4 Bb 16 Bb x Bb 1 BB , 2 Bb, 1 bb 11 22 Bb x bb 2 Bb, 2 bb 17 34 4 bb 8 32 Subtotals: 33 90 77 Grand total (#BB + #Bb +#bb): 200 Assume each pair mates and produces 4 offspring. Assume genetic make-up of offspring is as shown in second column. 50 mating pairs with 4 offspring = 200 individuals (400 alleles).

Calculate total number of b and B alleles from Table 1.2: # B alleles = (# BB individuals x 2) + (# Bb individuals) # B alleles = (33 x 2) + 90 = 156 # b alleles = (# bb individuals x 2) + (# Bb individuals) # b alleles = (77 x 2) + 90 = 244

Total # alleles in new population Table 1.3. Comparison of allele frequencies for original and new population Allele Total # alleles in new population Frequency of Alleles Population Original New B 156 p 0.400 b 244 q 0.600 Total: 400 1.000 # B alleles = (# BB individuals x 2) + (# Bb individuals) # B alleles = (33 x 2) + 90 = 156 # b alleles = (# bb individuals x 2) + (# Bb individuals) # b alleles = (77 x 2) + 90 = 244

Total # alleles in new population Table 1.3. Comparison of allele frequencies for original and new population Allele Total # alleles in new population Frequency of Alleles Population Original New B 156 p 0.400 0.390 b 244 q 0.600 0.610 Total: 400 1.000 Calculate frequency of each allele in the new population: p is the frequency of the B allele; q is the frequency of the b allele. Round to three decimal places, and record in Table 1.3 p = # B alleles/total # alleles; q = # b alleles/total # alleles. p = 156/400 = 0.390 q = 244/400 = 0.610

Pool class p and q data in Table 1.4. Table 1.4. Comparison of allele frequencies between original population and new population, for all tables   Frequency of Alleles p q Population Table Original New 1 0.4 0.6 2 3 Pool class p and q data in Table 1.4.

Activity 2 – Natural Selection Microevolution – Activity 2 – Natural Selection

Table 2.1. Results of natural selection on moths after three generations on group-assigned background Background color assigned to group Dark # Moths by color Generation 1 2 3 Initial Survivors Dark pink 16 Medium pink 48 Light pink 36 Total frequency 100 P 0.4 q 0.6 Create population of 100 moths and use background assigned to your table.

The captured moths are eaten and removed from the population. Table 2.1. Results of natural selection on moths after three generations on group-assigned background Background color assigned to group Dark # Moths by color Generation 1 2 3 Initial Survivors Dark pink 16 14 Medium pink 48 41 Light pink 36 29 Total frequency 100 P 0.4 q 0.6 One student will act as a bird feeding on the moths and try to catch them one at a time using tweezers for 30 sec. The captured moths are eaten and removed from the population. Count remaining live moths and record in survivors column.

Calculate the total number of generation 1 survivors, regardless of shade. Calculate the p and q values using the equations. p(frequency of B) = ((2 x #dark) + #medium) / total # of alleles q(frequency of B) = ((2 x #light) + #medium) / total # of alleles

Calculate initial numbers of dark, medium and light moths for 2nd gen using p and q calculated for survivors: # dark moths = p2 x 100 # medium moths = 2pq x 100 # light moths = q2 x 100

Repeat process of bird feeding on moths (2nd and 3rd generations). Remove and count remaining live moths, record “survivors” column for the 2nd generation.

(2 x #dark) + #medium)/total # of alleles Calculate the p and q values of the 3rd generation survivors as follows: p(frequency of B) = (2 x #dark) + #medium)/total # of alleles q(frequency of b) = (2 x #light) + #medium)/total # of alleles

Allele frequencies in survivors of each generation Pool all class data in Table 2.2. Table 2.2. Values of p and q over three generations, using three different backgrounds, by table   Allele frequencies in survivors of each generation Background  Table # 1 2 3 p q Dark Pink Medium Pink Light Pink

Weekly Data Sheet & Weekly Write-Up pages What’s Due Activity Weekly Data Sheet & Weekly Write-Up pages M1 497, 501, 505 M2 509, 513, 514 S3B 641, 645-646 Good luck on finals! Have a great winter break! PowerPoint available at: https://eeltown.org/evpp-110