Use the following information and the key to answer questions 17-19 Trees can ha

Question

Use the following information and the key to answer questions 17-19 Trees can have either scaly, rough or smooth bark. Trees homozygous for the 'S' allele of the bark texture gene have scaly bark, Ss heterozygotes have rough bark, while ss individuals have smooth bark. 1000 trees were scored for bark phenotype in a forest that was recently partially burned (ie. the population is not in Hardy-Weinberg equilibrium). 642 trees with scaly bark are observed, 348 trees have rough bark, and 10 have smooth bark a. 0.100 b. 0.184 c. 0.300 d. 0.816 Q17. What is the frequency of the S allele in this population? d Q18. What is the frequency of the s allele in this population? b Q19. Assuming random mating, what will be the expected frequency of heterozygotes in the next generation? c

Explanation / Answer

Q17. To estimate the frequency of the S allele, you have to consider that trees are di-allelic organisms, which means that they have two alleles for each characteristic (in this case the bark of the trees). So, in order to calculate the S allele frequency you need to determinate how many S alleles you have in your tree population as follows:

# of S alleles= ((# of S homozygous x 2) + # of Heterozygous (Ss)).

the homozygous are multiplied by 2 because they have 2 S alleles, the heterozygous are not multiplied because they only have one S allele.

# of S alleles= (642 x 2) + 348 = 1284 + 348= 1632.

Once the #of S alleles is determined you have to estimate the Frequency by dividing the previous result with the total number of alleles (1000x2), as follows:

Frequency of S allele= 1632/2000 = 0.816 (d).

Q18. In order to calculate the s allele frequency you need to determinate how many s alleles you have in your tree population as follows:

# of s alleles= ((# of s homozygous x 2) + # of Heterozygous (Ss)).

the homozygous are multiplied by 2 because they have 2 s alleles, the heterozygous are not multiplied because they only have one S allele.

# of S alleles= (10 x 2) + 348 = 20 + 348= 368.

Once the #of s alleles is determined you have to estimate the Frequency by dividing the previous result with the total number of alleles (1000x2), as follows:

Frequency of s allele= 368/2000 = 0.184 (b).

Q19. To calculate the heterozygous frequency (genotype frequency) you have to use the next formula:

2pq --> that it is employed to estimate the heterozygous genotype frequency (from the Hardy-Weinberg equation).

In the previous formula, p= frequency of the S allele and q= frequency of the s allele.

Heterozygous= 2(0.816 x 0.184) = 2(0.150144)= 0.300 (c).

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