Wild type yeast can synthesize all organic nutrients (amino acids, nucleotides,

Question

Wild type yeast can synthesize all organic nutrients (amino acids, nucleotides, vitamins, … etc.) needed to grow.  You have isolated 10 yeast mutants.  You determine that all mutants are defective for the organic nutrient known as “Z”.  The biosynthetic pathway required for synthesis of ”Z” is unknown, but you suspect that the organic chemicals A ,X ,H  and T are intermediates in the pathway.  You test each mutant for the ability to grow on minimal medium and on minimal media supplemented with each of the organic chemicals.  The results of your analysis are shown in the table above, where a minus sign (-) indicates no growth and a plus sign (+) indicates growth.  Use these results to answer the questions below.

1- Assuming that all the organic chemicals (A, X, H, and T) are in the Z synthesis pathway, which order is indicated by the data? Hint: A-H-T-X-Z indicates a metabolic pathway where A is converted to H which is converted to T which is converted to X which is converted to Z.

2- Which of the following mutants are blocked in the same step in the pathway?

3- You isolated a new recessive mutant called mutant 11.  Mutant 11 cannot grow on minimal medium.  Mutant 11 does not complement mutant 5.  Which statement below is correct if you take this information and the table provided with this problem into account?

Answers (b) and (c) are correct.

4- You mated mutant 4 to mutant 8 to create a diploid yeast strain. Assuming that the mutations in mutants 4 and 8 are recesssive, which of the following statements regarding the phenotype of this diploid strain is true?

A-T-H-X-Z A-H-T-X-Z A-X-H-T-Z Z-T-X-H-A A-T-X-H-Z Yeast Strain type of medium minimal minimalminimal+T minimal +X minimalminimal +Z wild type 4 10

Explanation / Answer

1) Order for the Z synthesis is A-X-H-T-Z

When minimal media supplemented with the T intermediate then many mutants showed growth which means, these mutants has the normal gene which converts T to Z thats why T is placed upstream of Z. Then many mutants which showed growth on media supplemented with H so it is placed upstream of T because mutants which showed growth on H have the normal copies of the genes which encodes an enzymes that converts H to T and T to Z. Next mutant 4 showed growth on the minimal media supplemented with X which finally converts X to H and H to T and finally T to Z and finally A must be present at the most upstream position.

Many mutants did not showed growth on the minimal media containing A intermediate which means many genes acts upstream of other intermediates but down stream of A and many gene mutants grows only on Z but not on other intermediates means that genes acts downstream of all the intermediates but upstream of Z.

2) 1 and 9 because mutants 1 and 9 did not showed growth on the minimal media supplemented with T (an intermediate present upstream of Z) means they had mutation in the gene which encodes for the enzyme that converts T to Z. Mutant 1 and 9 also not showed growth on minimal media supplemented with A, H and X because for the growth Z product is required and mutants lacks the normal copy of gene which converts T to Z . From the table we can conclude that mutant 1 and 9 have mutation in the same gene.

Whereas other mutants groups mentioned in the options do not acts on the same step of the pathway like 6 and 7 where mutant 6 can grow on the media supplemented with H and T whereas mutant 7 did not showed growth on any supplemented media beacuse it lacks an enzyme which converts T to Z (Final product of the pathway). Likewise others mutants given in the options do not block the same step.

3) Complementation only occurs when the mutation is in the different genes. Complementation test allows us to determine whether two mutation both of which produce similar phenotype are in the same gene i.e whether they are alleles or represent mutations in the separate genes. If mutant 11 and mutant 5 occurs due to mutation in same gene then both will not complement each other because neither of the mutant has the wild type copy of the gene so normal product of the gene is not produced.

If mutant 11 and mutant 5 has mutations in different genes example gene A is defective in mutant 11 and gene B is defective in mutant 5 then both will complement each other because mutant 11 provides the wild copy of gene B and mutant 5 provides the wild copy of gene A. So, the normal product of the genes A and B are present in the media and leads to the survival of both the mutants and in the question both the mutants 11 and 5 did not complement each other, so the answer will be Mutant 11 is defective in the same gene as in the mutant 5 or mutant 11 defective in the same step in the pathway as in the mutant 5.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.