Question

23 You continue your studies of snails. The woodland population from question 57 were moved to a grassland habitat. You relea57. Shell colour in snails is a single locus trait with co-dominant inheritance. The following frequencies of shell types occ

Question 57 is given below

23 You continue your studies of snails. The woodland population from question 57 were moved to a grassland habitat. You release 100 snails of each phenotype in an enclosed study area and measure survival after two weeks of bird predation #ofjuveniles at start on study (from Q57) 195 58. Phenotype survival after bird predation (%) 10 solid brown brown & green stripes solid green a) Es 60 409 50 198 timate the fitness of each phenotype and the mean fitness for the population in a grassland habitat. ) Estimate the expected frequency of each phenotype after one generation in the grassland. In calculating your answer, assume that surviving adults mate randomly and all phenotypes are equally fertile.
57. Shell colour in snails is a single locus trait with co-dominant inheritance. The following frequencies of shell types occur in a woodland population: solid brown 195 snails -409 snails - 198 snails brown & green stripes solid green What is the frequency of each allele in this population? a) How many individuals of each shell type are expected under Hardy-Weinberg b) equilibrium? c) Test the hypothesis that this snail population is in Hardy-Weinberg equilibrium for shell colour. d) What does your result suggest about the relative fitness of the three shell- colour phenotypes in woodland snails?
0 0
Add a comment Improve this question Transcribed image text
Answer #1

Fitness (w) of each phenotype is calculated by dividing each phenotype's survival rate by the highest survival rate among the 3 phenotypes.

(a) Fitness:

Solid-brown - 10/60 =0.167 (BB)

brown and green stripes - 60/60 = 1 (Bb)

Solid-green - 50/60 = 0.833 (bb)

Mean fitness = p2wBB + 2pqwBb + q2 wbb

Total individuals = 802

p (for BB) = 195/802 = 0.243

q (for bb) = 198/802 = 0.247

Mean fitness = 0.159

(b) Frequency for BB = pxp = 0.059

Frequency for bb = qxq = 0.061

Frequency for Bb = 2xpxq = 0.12

Add a comment
Know the answer?
Add Answer to:
Question 57 is given below 23 You continue your studies of snails. The woodland population from question 57 were moved...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT