In a particular species of guppy, tails can either be long or short and either feathered or straight. A cross between a true-breeding long, feather-tailed male guppy and a true-breeding short, straight-tailed female guppy produces progeny that all have short, straight tails.
a) express the genotypes of the adults in the cross and the genotypes of all the progeny.
b) determine the phenotypic frequencies of the progeny in an F2 generation.
c) a mating between a short feather-tailed female and a short, straight-tailed male from the F1 generation produces
30 short, straight-tailed guppies; 42 short, feather-tailed guppies; 10 long, straight-tailed guppies; and 14 long, feather-tailed guppies.
Analyze the data using the chi-squared fit test to determine if the frequencies of phenotypes are what would be expected from this cross© BrainMass Inc. brainmass.com October 24, 2018, 9:26 pm ad1c9bdddf
A cross between a true-breeding long, feather-tailed male guppy and a true-breeding short, straight-tailed female guppy produces progeny that all have short, straight tails.
We will designate a true-breeding long, feather-tailed male guppy as hhss.
We will designate a true-breeding short, straight-tailed female guppy as HHSS.
Capital letters will be used for dominant alleles. In this case, H = sHort tails, the dominant characteristic, whereas h = long tails. In addition, S = straight tails, and s = feathered tails.
We know these are the genotypes of the parents because all the F1 offspring are short and straight-tailed. Therefore, those two characteristics must be dominant.
Therefore, the cross is HHSS x hhss. All F1 must be HhSs -- without exception. That's why they all have short and straight tailed.
To determine the F2, we must set up a 4 x 4 grid.
HS Hs hS hs
HS HHSS HHSs HhSS HhSs
Hs HHSs HHss HhSs Hhss
hS HhSS HhSs hhSS hhSs
hs HhSs Hhss hhSs hhss
If you work out the phenotypes, you will end up with the classic 9:3:3:1 ratio for a regular complete dominant dihybrid cross. There's nothing unusual at this point. In other words, you'll end up with the following ratio:
9 short:straight, 3 ...
Hardy-Weinberg Equilibrium Question
Please help with the following problem.
If the frequency of the dominant phenotypes (AA + Aa) is 0.8 and the population is in Hardy-Weinberg equilibrium, can you calculate the allele frequencies? Can you calculate the genotype frequencies? Show your calculations.View Full Posting Details