15. Two petite yeasts, Mat-a and Mat-α are crossed. The diploid is grande. After a few mitotic divisions, the grande diploid is sporulated. The analysis of a large number of tetrads yielded a 2:2 ratio of petite: grande. A few potential scenarios describing the reason for this segregation pattern are stated below:
A. The parental strains had two different mitochondrial rho- mutations
B. One of the parents had a recessive nuclear petite mutation
C. Only one of the mitochondrial petite mutation is inherited in the tetrads
D. The mitochondria inherited are wild type.

Which one of the following options represents a combination of all correct statements?
(1) A, B and C         (2) B and D only
(3) A and D only      (4) B and C only

Introduction

Yeast genetics offers unique insights into mitochondrial and nuclear genome interactions, especially in studies of respiratory mutations. A classic problem involves crossing two petite yeast strains of different mating types and observing a 2:2 segregation of petite and grande colonies after meiosis. Understanding the origins of this pattern is crucial for advanced genetics learners aiming to master topics like mitochondrial inheritance, cytoplasmic versus nuclear petite mutations, and rho- yeast genetics.

Detailed Explanation of Each Option

Option A: Parental Strains Had Two Different Mitochondrial rho- Mutations

If both parents carry different mitochondrial ‘rho-‘ mutations (altered mitochondrial DNA), when crossed, recombination could restore functional mitochondrial DNA in the diploid. During sporulation, mitochondrial genomes might segregate so some spores inherit enough mitochondrial DNA to be respiratory-competent (grande) and others remain petite. This scenario can account for a 2:2 segregation if recombination events allow the restoration of respiratory function in half the spores.​

Option B: One Parent Had a Recessive Nuclear Petite Mutation

In yeast, “segregational” petites arise from recessive nuclear mutations. When a diploid with a heterozygous nuclear petite mutation is sporulated, Mendelian inheritance leads to a 2:2 segregation because the mutation resides in the nuclear genome. Two spores receive the mutation (petite), and two are wild type (grande). This aligns perfectly with observed 2:2 ratios in tetrad analysis.​

Option C: Only One of the Mitochondrial Petite Mutations Is Inherited in the Tetrads

Yeast typically exhibits cytoplasmic inheritance for mitochondrial traits. However, in some crosses (especially involving neutral or non-suppressive petites), only one parent’s mitochondrial DNA may be retained in each spore—leading to a 2:2 ratio if half the spores inherit wild-type mitochondria and half inherit petite mitochondria. This is a less common but possible explanation for the segregation observed.​

Option D: The Mitochondria Inherited Are Wild Type

If the mitochondria passed to the spores are always wild type (grande), all progeny would be grande, and a 2:2 segregation would not occur. Therefore, this explanation does not match the observed data and is incorrect for the scenario provided.​

Correct Combination(s) Table

Therefore, the correct answer is Option 1: A, B and C.

Summary Table: Correct Answer Options

SEO Note

This article covers yeast petite vs grande crosses, tetrad analysis segregation, and mitochondrial/nuclear petite mutations—essential concepts for CSIR NET genetics, molecular biology, and advanced life sciences preparation.

Correct answer: (1) A, B and C are the valid explanations for the 2:2 segregation of petite:grande yeasts in tetrad analysis.