Category: CSIR NET Life Science Previous Year Questions and Solution on EXTRA CHROMOSOMAL INHERITANCE

Category: CSIR NET Life Science Previous Year Questions and Solution on EXTRA CHROMOSOMAL INHERITANCE

CSIR NET Life Science Previous Year Questions and Solution on EXTRA CHROMOSOMAL INHERITANCE

16. Cytoplasmic male sterility (CMS) in plants is caused by mutation in the mitochondrial genome. CMS can be restored by a nuclear gene, restorer of fertility (Rf), which is a dominant character. If a male sterile pea plant is pollinated by a fertile male pea plant with Rf in heterozygous condition, the progeny obtained will have (1) all male sterile progeny (2) all fertile progeny (3) 50% of the progeny fertile and 50% male sterile (4) 75% of the progeny fertile and 25% male sterile

Inheritance of Cytoplasmic Male Sterility in Pea Plants

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

Yeast Petite and Grande Crosses

14. A mutant mating type mt strain of Chlamydomonas that was resistant to the antibiotic kanamycin (kanr) and herbicide PPT (pptr) was crossed to a wild type mating mt+ kams ppts strain that was sensitive to kanamycin and PPT. Twenty tetrads of the progeny were analyzed for mating type and resistance/sensitivity to kanamycin and PPT. The following observations were made: Type I Type II Type III mt kanr pptr mt kanr ppts mt kanr pptr mt kanr pptr mt kanr pptS mt kanr pptS Mt+ kanr ppts Mt+ kanr pptr Mt+ kanr pptr Mt+ kanr ppts Mt+ kanr pptr Mt+ kanr ppts Number of each type observed 8 9 3 The following statements were made to explain the observations: A. mt and ppt loci are on two different chromosomes B. Inheritance of mating type and ppt resistance/ demonstrating cytoplasmic sensitivity are inheritance C. Inheritance of kanamycin-resistance/sensitivity is demonstrating nuclear inheritance D. Nuclear inheritance is being demonstrated by mating type and ppt-resistance/sensitivity analysis Which one of the combinations of above statements is correct? (1) A and B (2) A and D (3) B and C (4) C and D

Chlamydomonas tetrad analysis – kanamycin resistance, PPT resistance and nuclear vs cytoplasmic inheritance

13. During development, many gene products are provided by the females to the eggs which are needed for normal development of the zygote. Such genes are called as maternal-effect genes. The following are a set of crosses between parents carrying a recessive mutant allele (m) and the offspring obtained: Phenotype of offspring Cross No. Genotype of parents Phenotype of offspring I m/+♂ X m/+♀ All normal II m/m♂ X m/+♀ All normal` III m/+♂ X m/m♀ All mutants IV m/m♂ X m/m♀ All mutants V m/+♂ X m/+♀ Both normal and mutant Which of the above cross(es) is/are indicative that the mutation is in a maternal-effect gene? (1) Cross III only (2) Cross V only (3) Cross I, II and III (4) Cross II and V

Maternal Effect Gene

12. In Drosophila, a cross was set between a male homozygous for alleles S+/S+ (phenotype A) and a female homozygous of S/S (phenotype B) ('S+' being a dominant allele and 's' a recessive allele). All of the F1 individuals thus obtained had the phenotype B. When F1 individuals were crossed among themselves, all progeny obtained were of phenotype A in F2. The following explanations were proposed for the results obtained: A. This is an example of cytoplasmic inheritance. B. This is exhibiting genetic maternal effect. C. This is a quantitative trait influenced by the environment. D. This is exhibiting gene interaction with E. The trait is showing position effect variegation. Which one of the following option is correct? (1) A only (2) B only (3) C only (4) D and E

Understanding Genetics Cross Result in Drosophila

11. A male snail homozygous for dextral alleles is crossed with a female homozygous for sinistral alleles. All the F1 individuals showed sinistral phenotype. When F1 progeny snails were self-fertilized all individuals of F2 progeny had dextral coiling. This experiment demonstrated (1) dominant epistasis as dextral allele is dominant over sinistral allele (2) recessive epistasis as in F2 dextral allele appeared in homozygous condition (3) maternal effect as the nuclear genotype of the F1 mother has governed the phenotype of the F2 individuals. (4) maternal inheritance as the mitochondrial genes of the F1 mother has governed the phenotype of the F2 individuals.

Maternal Effect in Shell Coiling of Snails

10. Maternal inheritance of coiling of shell in snail (limnaea peregra) is well established. The dextral coiling depends on dominant allele D and sinistral coiling depends upon recessive allele d. A female F1 progeny of dextral(Dd) type is crossed with a male sinistral snail. What will be the ratio of heterozygous: homozygous individuals in its F2 progeny? (1) 3:1 (2) 1:1 (3) 1:3 (4) 1:2:1

Maternal Inheritance of Coiling of Shell in Snails

9. In Neurospora, the mutant stp exhibits erratic stop-and-start growth. When a female of stp strain is crossed with a normal strain acting as a male, all progeny individuals showed stp mutant phenotype. However, the reciprocal cross resulted in all normal progeny, individuals. These results can be explanined on the basis of A. maternal inheritance B. sex limited inheritance C. sex influenced inheritance D. stp mutation may be located in mitochondrial DNA The most appropriate statement or combination of the above statements for explaining the experimental results is: (1) A and C (2) C only (3) A and D (4) B and D

Understanding the Inheritance of the stp Mutant Phenotype in Neurospora

8. In Neurospora, the mutant stp exhibits erratic stop-start growth. The mutant site is known to be in the mitochondrial DNA. If an stp strain is used as the female parent in a cross to a normal strain acting, as the male, what type of progeny can be expected? (1) All Start and stop mutant (2) All wild type (3) Majority of Start and stop mutant (4) Majority of wild type

Maternal Mitochondrial Inheritance of stp Mutant in Neurospora

7. The characteristic of mitochondrial genome is? (1) Intron free DNA (2) Repetitive DNA (3) Polycistronic transcription (4) Satellite DNA

Characteristics of Mitochondrial Genome Explained

6. A poky Neurospora was crossed with normal Neurospora and following results were obtained ♀ Poky X ♂ Normal  all poky ♀Normal X ♂ Poky  all Normal The mode of inheritance is (1) Maternal Inheritance (2) Maternal effect (3) X-Linked (4) Sex influenced

Understanding the Mode of Inheritance in Poky Neurospora Crosses

5. Yeast with petite colony when crossed with wild type generates no petite colony. The most probable mode of inheritance is (1) Chloroplast (2) Mitochondria (3) Episomal (4) Nuclear

Mode of Inheritance of Petite Colony in Yeast

4. Gene for fungal resistance in certain organism resides in chloroplast. If a susceptible female and resistant male are crossed, the progeny will be (1) All resistance (2) All susceptible (3) Half resistance and half susceptible (4) Cannot be predicted

Gene for Fungal Resistance in Chloroplast

3. Extra-chromosomal inheritance can be detected by- (1) Back cross (2) Test cross (3) Reciprocal cross (4) Dihybrid cross

Detecting Extra-Chromosomal Inheritance

2. Reciprocal crosses yield different result in (1) Mendelian inheritance (2) Nuclear inheritance (3) Cytoplasmic inheritance (4) Polygenic inheritance

Understanding Reciprocal Crosses and Their Different Outcomes in Genetic Inheritance

Cytoplasmic Male Sterility in maize is due to- (1) Cytoplasm (2) Plastid (3) Mitochondria (4) Ovary

Role of Mitochondria and Mechanism

Latest Courses

CSIR UGC NET Life Science Course

CSIR UGC NET Life Science Course

Let's Talk Academy is the number one CSIR NET Life Science coaching in Jaipur for Life Science. At Let's Talk Academy, you can be sure to get the right coaching strictly in adherence to the CSIR NET Life Science course.  CSIR NET Life Science is an exam that requires in-depth knowledge, consistent practice, and the right guidance to crack. If you are looking for a place where you can get the best CSIR NET Life Science coaching, you are at the right place.

ICMR JRF Life Science

ICMR JRF Life Science

If you are serious about cracking the ICMR Life Science JRF exam, Let’s Talk Academy is your ultimate destination. Join now and take the first step toward a successful research career!

IIT JAM / CUET- PG

IIT JAM / CUET- PG

Are you aspiring to crack the IIT JAM Life Sciences exam with flying colors? Choosing the right coaching institute is crucial for your success. Let's Talk Academy (LTA) has established itself as the best coaching institute for IIT JAM Life Sciences, helping students achieve their dreams through expert guidance, structured study material, and personalized mentorship.