A truebreeding C.elegans with blisters (bli-6/bli-6) and normal length (lon-1^+/
ID: 168032 • Letter: A
Question
A truebreeding C.elegans with blisters (bli-6/bli-6) and normal length (lon-1^+/lon-1^+) was crossed to a male with no blisters (bli-6^+/bli-6^+) and a long body (lon-1/lon-1). Both of these genes are autosomal. All of the F1 appeared wild type. a. A testcross was done with an F1 female and a tester male. The following data was generated from the testcross. Fill in the table below. b. Without calculating the recombination frequency, can you determine if the 2 genes are linked or independently assorting? Explain. c. Calculate the map distance between the 2 genes. d. If C.elegans is 2n=12, draw a meiocyte at Metaphase I from an F1 hermaphrodite. Be sure to include bli-6 and lon-1 (and the appropriate alleles) on your map.Explanation / Answer
Answer:
a).
Phenotype of progeny
Partental or recombinant
# observed
Blisters and long
Recombinant
61
No Blisters and normal length
Recombinant
58
Blisters and normal length
Parental
255
No Blisters and long body
Parental
249
b). Yes, we can determine whether the genes are linked or independently assorting by using number of progeny without recombination frequency. In test cross, all progeny are produced in equal proportion when the genes assort independently. If the genes are linked, the test cross progeny would not have an equal proportions.
c).
Recombination frequency (%) = Map distance (map units)
Recombination frequency = (no of recombinant/Total progeny)*100
RF = (119/623)100 = 19.1%.
The distance between the genes is 19.1 map units
Phenotype of progeny
Partental or recombinant
# observed
Blisters and long
Recombinant
61
No Blisters and normal length
Recombinant
58
Blisters and normal length
Parental
255
No Blisters and long body
Parental
249
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.