1. The genotypes of four strains of Neurospora are unknown. However, we do know

Question

1. The genotypes of four strains of Neurospora are unknown. However, we do know that all strains require arginine and have the growth characteristics shown below.

Nutrition and Accumulation

Growth on
Strain Minimal Ornithine Citrulline Arginine

1 - - + +

2 - - - +

3 - - - +

4 - - - +

Pairwise complementation tests of the four strains:

1

1 0 2

2 + 0 3

3 0 0 0 4

4 + 0 0 0

+ = heterokaryon grows on minimal media

0 = no growth of heterokaryon on minimal media

Crosses among mutants yielded prototrophs in the following percentages:

1 x 2 25%

1 x 4 25%
2 x 4 0.004%

1 x 3 none detected among 1 million ascospores

2 x 3 0.002%
3 x 4 none detected among 1 million ascospores

A. How many distinct mutational sites are represented among these four strains affecting how many polypeptide chains?

B. Give the genotypes of the four strains, using a consistent and informative set of symbols.

C. Give the map distances between all pairs of linked mutations.

D. What percentage of prototrophs would be expected among ascospores of the following types:

Strain 1 x wild type Strain 3 x wild type Strain 2 x wild type Strain 4 x wild type

Explanation / Answer

A).

Nutrition and accumulation data can be used to infer where each mutant is blocked.

1 accumulates ornithine, and it grows only if citrulline or arginine is added. Hence, 1 is blocked in the conversion of ornithine to citrulline.

2 and 3 are blocked in the conversion of citrulline to arginine.

Mutant 4 is more complex. It grows only if supplemented with arginine, but accumulates ornithine.

4 must be a double mutant and is blocked in the conversion of ornithine to citrulline, as well as from citrulline to arginine.

B).

Genotype of 1 - CCOO

Genotype of 2 and 3 - CCoo

Genotype of 4 - ccOO

Where,

C = citruline

O = ornithine

C).

Map distance is 25 mapunits between 1 and 2.

3 and 4 may be on different strands of chromosome.

D).

The percentage of prototrophs expected from strain 1 x wild type = 100%

The percentage of prototrophs expected from strain 2 x wild type and strain 3 x wild type = 50%

The percentage of prototrophs expected from strain 4 x wild type = 50%

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