In Drosophila, balancer chromosomes are used to keep all the alleles on one chromosome together. A balancer contains multiple inversions; so that when it recombines with the corresponding wild type chromosome, no viable cross over products are formed. Balancers also carry an allele for a dominant phenotype.

A Drosophila male with sepia eye color is crossed to a female carrying a third chromosome balancer (TM6B). The allele for sepia phenotype (se) is located on chromosome 3 and is recessive to the wild type eye color. The dominant marker for TM6B is a tubby phenotype. Further, an individual homozygous for TM6B balancer does not survive. F1 progeny with tubby phenotype is se/+.

The F₂ progeny is expected to have:

Correct option is C

Explanation-

A Drosophila male with sepia eye color (se) (a recessive mutation on chromosome 3) is crossed with a female carrying a balancer chromosome TM6B.
TM6B carries:
          1.   Multiple inversions (which prevent recombination)
          2.   A dominant tubby (Tb) phenotype
TM6B homozygotes are not viable, meaning only heterozygous flies with TM6B can survive.
The F₁ progeny with tubby phenotype (i.e., carrying TM6B) are sib-mated to produce F₂ generation.

Genotypes Involved:
Sepia eye color (se) is recessive; wild-type is dominant.
TM6B = a balancer with tubby marker and lethal when homozygous.
Initial cross:
          Male: se/se
          Female: +/TM6B (wild type at se locus)
So the F₁ progeny will be:
          se/+ (wild-type eye), and
          se/TM6B (tubby due to TM6B, but heterozygous and viable)
          Only se/TM6B tubby flies are selected and sib-mated for the F₂ generation.

F₁ Tubby × F₁ Tubby Cross (se/TM6B × se/TM6B):
Using a Punnett square:​
                                                                                    se                                                         TM6B
se                                                                              se/se                                                    se/TM6B
TM6B                                                                    se/TM6B                                             TM6B/TM6B  (lethal)
So the F₂ genotypes:
se/se → sepia eye, non-tubby
se/TM6B → wild-type eye, tubby
TM6B/TM6B → lethal (dies)

F₂ Phenotypic Ratio:
se/se (sepia) = 1
se/TM6B (tubby) = 2
TM6B/TM6B = 1 (but lethal, so not counted)
So surviving ratio is:
sepia : tubby = 1 : 2

 Correct Option is c -  sepia and tubby flies in a ratio of 1:2