love and dowe (1993) studied the nesting results of a whip tail lizard from New
ID: 262941 • Letter: L
Question
love and dowe (1993) studied the nesting results of a whip tail lizard from New Mexico and noticed two color morphs in an interbreeding population, whip tails with white stripes and whip tails with gray stripes. DNA sequencing showed them to be the same species, Cnemidophorus inornatus. A population of neonates from these lizards was censused at Red Bluff Reservoir and found to contain 260 white-striped baby whip tails and 430 gray-striped baby whip tails. Assume striping color is controlled by a single locus. a.) First, calculate the frequencies of all three possible genotypes by assuming white is the dominant striping pattern and that the population is in Hardy-Weinberg Equilibrium. Repeat the calculations by then assuming gray is the dominant color and the population is in HWE. Gray recessive = gg, White dominant = GG: GG = ____________________ Gg = ____________________ gg = _____________________ White recessive = ww, Gray dominant = WW WW = ___________________ Ww = ___________________ ww = ___________________
b.) This population is found in a hillside made up of gray basalt. Assuming the gray morph is dominant, empirical evidence suggests the white stripe morph has a greater mortality rate than the gray striped morph since it is easier for hawks flying overhead to see the white morph against the volcanic rock. If 303 gray-striped whip tails survived to adulthood but only 150 white-striped whip tails did, calculate the fitness of the white-striped phenotype relative to the gray one. Fitness of white stripe compared to gray stripe* = *Survival rate of gray striped: (# gray survivors)/(# of original gray pop) = ___ Survival rate of white striped: (# white survivors)/(# of original white pop) = ___ Now, normalize one allele by setting its value to 1 and consider the heterozygote having the same fitness as the homozygote dominant condition: ?WGG: (survival rate of gray) / (survival rate of gray) = 1.0 ?WGg: (survival rate of gray) / (survival rate of gray) = 1.0 ?Wgg: (survival rate of white) / (survival rate of gray) = _____
c.) Now calculate mean fitness (‘w-bar’) and use that to predict the effect of selection on the next generation for all three genotypes. p 2WAA + 2pqWAa + q2Waa = mean W:
d.) Suppose Wright and Lowe return 4 years later and find that 69.2% of the population is composed of white striped morphs. Propose scenarios (plausible ones, please) to account for this
Explanation / Answer
a)
*white dominant (GG) gray recessive (gg)
GG and Gg = 260 gg = 430;
frequency of gg(q2) = 430/690 = 0.62 ; q2 = 0.62, q = 0.78. therefore, p = 0.22. So, p2 = 0.04. Therefore, 2pq = 0.18
GG = 0.04, Gg = 0.18, gg = 0.62
*gray dominant (WW) white recessive (ww)
WW and Ww = 430, ww = 260;
ww = q2 = 260/690 = 0.37, q = 0.60 ; p = 0.40 ; p2 = 0.16 ; 2pq = 1 - (0.37+0.16) = 0.47
WW = 0.16 ; Ww = 0.47 ; ww = 0.37
b)
Survival rate of gray = 303/430 = 0.70 ; Survival rate of white = 150/260 = 0.57
Relative survival rate of white to gray = 0.57/0.70 = 0.714
c)
mean fitness = 0.57 + 0.70 / 2 = 1.12/2 = 0.56
The next generation will have similar levels of fitness. fitness of dominant homozygote and heterozygote are relatively same.
d)
Majority of the population could be comprised of white tailed lizards due to genetic drift. One plausible scenario is that over 4 years vegetation grew around the area which removed the natural fitness in gray lizards as gray contrasting background was no longer present exposing both phenotypes to predators equally. futhermore, white tailed could be innately more successful at reproduction thus higher numbers of white tailed lizards than gray ones when advantage of camoflage got removed.
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