Question

Extra Problem 2. Four mutant strains of Neurospora (#1-4) were examined by complementation testing. Results are...

Extra Problem 2. Four mutant strains of Neurospora (#1-4) were examined by complementation testing. Results are shown below (+ indicates growth; - indicates no growth).

1

2

3

4

1

-

+

-

+

2

+

-

+

-

3

-

+

-

+

4

+

-

+

-

Crosses among the mutant strains yielded prototrophs in the following percentages:

            1 x 2   25.0%                        2 x 3   25.0%                        3 x 4   25.0%

            1 x 3   0.02%                        2 x 4   none

A. How many genes are mutant among these strains?

B. Give map distances between all pairs of linked mutations and indicate which mutations are unlinked.

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Answer A- There are two mutant genes present in these strains. The mutant groups are group1=1,3 (because neither the complementation of 1 or 3 could rescue growth of 1), group2=2,4 (because neither the complementation of 2 or 4 could rescue growth of 4).

Answer B-

The mutations 1 and 3 are linked. Similarly mutations 2 and 4 are also linked

The recombinant frequency of (1-2)=25% (unlinked), (1-3)=~0% (linked), (2-3)=25% (unlinked), (2-4)=0% (linked) and (3-4)=25%(unlinked).

The gene map can be drawn as follows,

Mutations 1 and 2 are unlinked because 1 and 2 assort independently. Similarly mutations 3 and 4 are unlinked because 3 and 4 assort independently.

Add a comment
Know the answer?
Add Answer to:
Extra Problem 2. Four mutant strains of Neurospora (#1-4) were examined by complementation testing. Results are...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • Question 9 In an analysis of other rII mutants, complementation testing yielded the following results: Mutants      Res...

    Question 9 In an analysis of other rII mutants, complementation testing yielded the following results: Mutants      Results (+/- lysis) 1,2                 -- 1,3                 -- 1,4                 + 1,5                 + Which mutants are in the same complementation group as mutation #1? a. mutant 2 b. mutant 3 c. mutant 4 d. mutant 5 e. None of these mutations are in the same complementation group. f. All of these mutations are in the same complementation group.

  • Four auxotrophic mutant strains of E. coli (1, 2, 3, 4) are each blocked at a...

    Four auxotrophic mutant strains of E. coli (1, 2, 3, 4) are each blocked at a different step of the same metabolic pathway. This pathway involves compounds L, M, N, O, and P. Each of these compounds is tested for its ability to support growth of mutant strains, with the results given below. (+= growth, __ = no growth) Compound Used as a Supplement L M N O P Mutant 1 + __ + + __ Mutant 2 __ __...

  • 14..........In Neurospora (a fungus), a complementation test was performed on tive haploid mutants Wild-type in appearance,...

    14..........In Neurospora (a fungus), a complementation test was performed on tive haploid mutants Wild-type in appearance, and means showed abnormal branching. (3 pts) It the table below, complementation crosses were made to establish which mutants rescue the mutation. A complementation group will contain mutants that can rescue the defective phenotype. OWN These results indicate that the mutations were in A) 1 gene. B) 2 genes. C) 3 genes. D) 4 genes. The polarity of DNA synthesis is: E) 5 genes....

  • Eight mutant bacteriophage strains cannot lyse a certain type of bacteria that can be lysed by...

    Eight mutant bacteriophage strains cannot lyse a certain type of bacteria that can be lysed by wild-type bacteriophages. The mutant strains were allowed to infect the bacteria in a complementation test. A "+" indicates that lysis occurred with coinfection. A "-" indicates that lysis did not occur. 1 2 3 4 5 6 7 8 1 - + + + + - + + 2 - + + - + - + 3 - - + + + - 4...

  • 2) Eight different point mutations are studied by complementation test and by der mapping. The results...

    2) Eight different point mutations are studied by complementation test and by der mapping. The results of all pair-wise complementation tests between shown below. A) Classify these eight mutants into genes. 1 2 3 4 B) The eight point mutations are crossed with three deletion mutants. Taking these and the complementation data into account, indicate the order of mutations 1-8. Draw circles around paired mutations in the same gene. Also draw three lines beneath the map to indicate the position...

  • Can you please help me with this problem: 1. Four different uracil auxotrophs of Neurospora, a...

    Can you please help me with this problem: 1. Four different uracil auxotrophs of Neurospora, a eukaryotic mold, are tested for growth on uracil and uracil precursors. The data are shown in the following table. A plus sign (+) means growth. Compound A ІВ IC ID Uracil Mutant 1 + + + Mutant 2 + Mutant 3 + + + Mutant 4 + + + Diagram the uracil pathway, showing the step at which each mutant is blocked. (4 pts)

  • please help NOT QUESTION 2 NOT QUESTION 4 1. The growth responses in the following chart...

    please help NOT QUESTION 2 NOT QUESTION 4 1. The growth responses in the following chart were obtained by growing four mutant strains of Neurospora on different media, each containing one of our related compounds. A B C and all of which are precursors to product C. All mutants require product C for growth Precursor/Product B Mucation A cate which shep Show a simple near biosynthetic pathway of the four precursors and the end product blocked by each of the...

  • 6. Drosophila have antennae on their heads. Two mutants with abnormal antennae were obtained in a...

    6. Drosophila have antennae on their heads. Two mutants with abnormal antennae were obtained in a mutant hunt. After establishing true breeding strains of the mutants, Sally did the following crosses. Draw conclusions from her results. Suggestion: where helpful, outline the cross as a Mendelian set-up, choose letters. a. ant 1 X wild type, produced all flies with normal antennae. (2 pts) b. ant 2 X wild type produced all flies with normal antennae. (2 pts) Next Sally crossed the...

  • B12030 (12) Defining the gene E cow mutants are isolated that cannot utilize tetrahydro-cannabinol (THC) as...

    B12030 (12) Defining the gene E cow mutants are isolated that cannot utilize tetrahydro-cannabinol (THC) as a source of carton. energy or anything else. The map positions in minutes of the the mutations are 18 minutes #2.4 5.0.7. 10, 11, 12, 13]. [42 minutes. M 1: 151 minutes: 391: 166 minutes: #8: (85 minutes: #141 fa) What is the minimum number of genes involved in THC utilization? OVA. AM 2 5 6 7 10 11 12 13 4 0 0...

  • b) (2 points) Describe the experiment and the results that allow her to conclude that the...

    b) (2 points) Describe the experiment and the results that allow her to conclude that the five mutants result from mutations in five different genes. 097 m iii) Cysteine, cystathionine and homocysteine are precursors to methionine. She grew all five mutant lines on minimal medium with either cysteine, cystathionine, homocysteine or methionine, with the following results (+ indicates growth of the mutant line, - indicates no growth): + + + + + + Cysteine cystathionine homocysteine methionine met-1 met-2 met-3...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT