Amale fly that is homozygous for recessive alleles at two linked autosomal loci is mated with a wild-type female fly that is homozygous dominant for both alleles. all f1 offspring are then randomly mated. if the loci are so close as to have no recombination, what proportion of offspring will be homozygous for both mutations in the f2 generation?

In F2 generation, 50% of the offspring will be homozygous for both mutations, which will be 25% homozygous for dominant alleles, and 25% for recessive alleles.

Explanation:

Both loci are found in an autosomal chromosome, so genotypic frequencies will be independent of the sex of the parents or the progeny. Also, both loci are linked, so there is no recombination, and they segregate always together.  

Gene 1 will be M (M for dominant, and m for recessive), and gene 2 will be P (P for dominant, and p for recessive). So, if a double recessive male (m/p;m/p) is mated with a double dominant female (M/P; M/P), it will produce a progeny 100% heterozygous (M/P;m/p), as in the attached Punnett Square.

Having into account that there is no recombination, alleles always will be segregated together (M/P: they always will be together as well as m/p).

Therefore, F2 will be like is seen in the Punnett Square, in which 50% of progeny is homozygous for both genes: 25% for dominant alleles and 25% for recessive alleles.

thats not even an aswer choice

my teacher would say no s