Page2 Biology 311 Midterm Exam- Fall 2015 Question 1 13 marks) breeding plants t
ID: 88857 • Letter: P
Question
Page2 Biology 311 Midterm Exam- Fall 2015 Question 1 13 marks) breeding plants that were either wild. Three different crosses were performed with different linked genes a,b, c) such that F1 type (+)or mutant (lowercase letter for three genes. A test cross was performed and were for each of the three recorded. the number of plants in each phenotypic class was observed in 3-point 30 317 63 10 339 28 107 142 124 1265 a) For each the crosses (1.2 and 3), what were the genotypes of the pure breeding of parents? (3 marks) Look at each cross independently- they may show different gene orders.3marks) bacExplanation / Answer
1) a) The phenotypes with the largest number of offsprings are the pure breeding parental types. Therefore, the genotype of pure breeding parents are,
data set 1 ---> +/+, +/+, +/+ and a/a, b/b, c/c
data set 2 ---> +/+, b/b, +/+ and a/a, +/+, c/c
data set 3 ---> +/+, +/+, +/+ and a/a, b/b, c/c
b) The least number of offsprings represent the double cross over types. By comparing the double cross over types with parental types, gene order can be found. From the data sets, the double cross over types can be found. By comparing this with the parental types, gene order can be predicted without doing any calculation. By this method, the gene order for data 1, 2 and 3 are predicted and found to be bac, bac and acb respectively.
c) For cross 2, the gene order is bac.
The distance between a and b is (30 + 3 + 6 + 34) / 982 = 73/982 = 0.074 = 7.4%
The distance between a and c is (6 + 137 + 142 + 3) / 982 = 288/982 = 0.293 = 29.3%
The distance between b and c is (30 + 137 + 142 + 34) / 982 = 343/982 = 0.349 = 34.9%
The linkage map is ,
c 29.3 a 7.4 b
l------------------------------I----------------------I
34.9
d) If we take double cross overs also into account, the distance between b and c is
(30+137+142+34+3+3+6+6) / 982 = 361/982 = 0.367 = 36.7%
e) The calculated recombination frequency will be underestimate since the double cross overs are not accounted.
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