Question

1. You are given the task of creating a genetic map of three genes of Drosophila melanogaster . The genes are all located on the same chromosome. The genes are for body color, eye color and antenna formation. The alleles for body color are yellow and gray (wild-type), the alleles for eye color are sepia and red (wild-type) and the alleles for antenna are short and normal (wild-type). The mutant alleles for body and eye color are recessive, while the mutant allele for antenna formation is dominant. You follow the procedure for determining the order and map distance for the genes and obtain the following data:

Short, gray, sepia = 240 Short, yellow, red = 2 Short, yellow, sepia = 500 Short, gray, red = 22 Normal, yellow, red = 270 Normal, gray, sepia = 5 Normal, yellow, sepia = 15 Normal, gray, red = 444

Hints for drawings: size matters! Don’t draw all genes equidistant. If two genes are further away from each other than another set, make sure to draw it that way.

Based on the data given:

1. Define each gene and allele. Based on how you defined the alleles, what was the genotype and phenotype of the P gen? Make sure to notate alleles properly and to draw the chromosomes correctly. Make sure the genes are in the correct order.

2. What was the genotype and phenotype of the F1 gen? Make sure to notate alleles properly and to draw the chromosomes correctly. Make sure the genes are in the correct order.

3. What was the genotype and phenotype of the Test cross that was performed? Make sure to notate alleles properly and to draw the chromosomes correctly. Make sure the genes are in the correct order.

4. What are all of the map distances between the genes given? Make sure to draw the chromosomes correctly, the genes are in the proper order and units are used.

Answer 1

Answer:

a).

Short (T) vs normal antennale (t)

Gray (G) vs yellow (g)

Red (R) vs sepia (r)

Parent 1 = short antennae, yellow body, sepia eyes = Tgr/Tgr

Parent 2 =wild-type antennae, gray body, red eyes = tGR/tGR

b).

Tgr/Tgr x tGR/tGR ---Parents

Tgr/tGR ------------------F1

F1 genotype = Tgr/tGR

F1 phenotype = Wild-type antennae, gray body, red eyes

c).

Tester that used to cross with F1 during testcross = tgr/tgr (wild-type antennae, yellow body, sepia eyes)

d).

Order of genes = t-r-g

Gene map = t----------2.94mu--------r----------34.51mu--------------g

Explanation:

Short, gray, sepia = 240 (TGr)

Short, yellow, red = 2 (TgR)

Short, yellow, sepia = 500 (Tgr)

Short, gray, red = 22 (TGR)

Normal, yellow, red = 270 (tgR)

Normal, gray, sepia = 5 (tGr)

Normal, yellow, sepia = 15 (tgr)

Normal, gray, red = 444 (tGR)

Hint: Always non-recombinant genotypes are large numbered than the recombinant genotypes.

Hence, the parental (non-recombinant) triple heterozygous genotypes is Tgr/tGR

1).

If single crossover occurs between t&g..

Normal combination: Tg/tG

After crossover: TG/tg

TG progeny= 240+22 = 262

tg progeny = 270+15 = 285

Total this progeny = 547

Total progeny = 1498

The recombination frequency between t&g = (number of recombinants/Total progeny) 100

RF = (547/1498)100 = 36.51%

2).

If single crossover occurs between g&r..

Normal combination: gr/GR

After crossover: gR/Gr

gR progeny= 2+270 = 272

Gr progeny = 240+5 = 245

Total this progeny = 517

The recombination frequency between g&r = (number of recombinants/Total progeny) 100

RF = (517/1498)100 = 34.51%

3).                                        

If single crossover occurs between t&r..

Normal combination: Tr/tR

After crossover: TR/tr

TR progeny= 2+22 = 24

tr progeny = 5+15 = 20

Total this progeny = 44

Total progeny = 1498

The recombination frequency between t&r = (number of recombinants/Total progeny) 100

RF = (44/1498)100 = 2.94%

Recombination frequency (%) = Distance between the genes (mu)

Order of genes = t-r-g

Gene map = t----------2.94mu--------r----------34.51mu--------------g