In Drosophila melanogaster, vestigial wings (vg) is recessive to normal wings (vg+), black body (b) is recessive to gray body (b+), and purple eyes (pr) are recessive to red eyes (pr+). All three genes are located on the same chromosome. A female fly with vestigial wings, black homes body, and purple eyes is crossed to a male with normal wings, gray body, and red eyes. The F1 females produced by this cross were mated with vestigial, black, purple males in a test cross. After doing a test cross the following pheonotypes were obtained. 1779 vestigial wings, an ebony body and purple eyes
1665 normal wings, a brown body and red eyes
252 normal wings, an ebony body and purple eyes 241 vestigial wings, a brown body and red eyes
131 normal wings, an ebony body and red eyes
118 vestigial wings, a brown body and purple eyes
13 vestigial wings, an ebony body and red eyes
9 normal wings, a brown body and purple eyes.
1. Determine the order of these genes on the chromosome.
2. Calculate the map distances between the genes. Construct a genetic map for these three loci.

Correct progeny phenotype:

Explanation:

We know that

•Normal wings expressed by vg+ is dominant over vestigial wings, vg

•Gray body b+ is dominant over black body

•Red eyes, pr+, is dominant over purple ayes, pr

We have the number of descendants of each phenotype product of the tri-hybrid cross.

•1779 vestigial wings, black body and purple eyes vg b pr  

•1665 normal wings, a grey body and red eyes vg+ b+ pr+

•252 normal wings, an black body and purple eyes vg+ b pr

•241 vestigial wings, a gray body and red eyes vg b+ pr+

•131 normal wings, an black body and red eyes vg +b pr+

• 118 vestigial wings, a gray body and purple eyes vg b+ pr

•13 vestigial wings, an black body and red eyes vg b pr+

• 9 normal wings, a gray body and purple eyes vg+ b+ pr

The total number of individuals is 4208.

In a tri-hybrid cross, it can occur that the three genes assort independently or that two of them are linked and the third not, or that the three genes are linked. In this example, in particular, the three genes are linked on the same chromosome.

Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the genotypes of the parental gametes with the ones of the double recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:

Parental)

Double recombinant)

Simple recombinant)

Comparing parental with the double recombinants we will realize that between  

and  

They only change in the position of the alleles pr/pr+. This suggests that the position of the gene pr is in the middle of the other two genes, vg and b, because in a double recombinant only the central gene changes position in the chromatid.  

So, the order of the genes is:

vg pr b

Now we will call Region I to the area between vg and pr and Region II to the area between pr and b.

Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between vg and pr genes, and P2 to the recombination frequency between pr and b.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of simple recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals.  

So:

Parental)

•1779 vestigial wings, black body and purple eyes vg pr b  

•1665 normal wings, a grey body and red eyes vg+ pr+ b+  

Double recombinant)

• 13  vg pr+ b  

• 9  vg+ pr b+  

Simple recombinant)

• 252  vg+ pr b  

• 241 vg pr+ b+  

• 131  vg+ pr+ b  

• 118  vg pr b+  

P1 = (R + DR) / N

P1 = (252+241+13+9)/4208

P1 = 515/4208

P1 = 0.122

P2= (R + DR) / N

P2 = (131+118+13+9)/4208

P2 = 271/4208

P2 = 0.064

Now, to calculate the recombination frequency between the two extreme genes, vg and b, we can just perform addition or a sum:

P1 + P2= Pt

0.122 + 0.064 = Pt

0.186=Pt

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).  

The map unit is the distance between the pair of genes. Every 100 meiotic products, one of them results in a recombinant product. Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:

GD1= P1 x 100 = 0.122 x 100 = 12.2 MU

GD2= P2 x 100 = 0.064 x 100 = 6.4 MU

GD3=Pt x 100 = 0.186 x 100 = 18.6 MU

vg pr b

                    R1                                 R2

vg12.2MUpr—

                                   pr6.4 MUb—

vg 18.6 MUb

label a cell membrane

label b cytoplasm

label c ribosomes

label d dna

the membrane forms the outside of the cell

cytoplasm is the liquid within

ribosomes are the components within that distribute rna

the dna is the genetic make-up

the answer is c. it will give up it's valence electron to the other atom.