In Drosophila, sepia eyes (se), curled wings (cu) and ebony body (e) are found, in this order, on chromosome 3. Sepia is 22 map units from curled, which is 12 map units from ebony. A true-breeding female with sepia eyes and ebony body is crossed to a true- breeding male who has curled wings. The resulting F1 progeny from this cross are all wild type. An F1 female is test crossed to a male expressing all three traits. Assuming a Coefficient of coincidence of 1: What proportion of the F2 progeny will express only the sepia trait?  If there are 1000 total flies, what number will express curved only?

A PORTION OF THE LINKAGE MAP OF CHROMOSOME 2 IN THE TOMATO IS ILLUSTRATED HERE. ci (compound influorescence) o (oblate) - 15 CM 20 CM p (peach) THE OBLATE PHENOTYPE IS A FLATTENED FRUIT, THE PEACH PHENOTYPE IS HAIRY FRUIT (LIKE A PEACH), AND COMPOUND INFLORESCENCE MEANS CLUSTERED FLOWERS. IGNORE THE PEACH LOCUS. AMONG 1000 GAMETES PRODUCED BY A PLANT OF GENOTYPE O CI /+ +, WHAT TYPES OF GAMETES WOULD BE EXPECTED, AND WHAT NUMBER WOULD BE EXPECTED OF EACH?

Two genes, m (apical meristem) and r (starburst rhizosomes), are linked on chromosome #5 in pea plants. Both genes are recessive to wild type and found to be 8 MU apart. If a wild type heterozygote is test- crossed, and produces 91 apical meristem plants and 8 apical, starburst plants from a total of 200 plants were produced, how many wild type plants will be produced? 4 8. 12 16

In tomatoes, regular leaves (L) are multilobed and serrated and potato leaves (l) are broad, smooth, and single (Image 1). Red fruit (F) is dominant to yellow fruit (f). A cross is carried out between two pure lines of tomato plants, one having regular leaves and red fruit and the other having potato leaves and yellow fruit. The F1 generation all have regular leaves and red fruit. The F1individuals are then crossed with one another. The information below represents two sets of data collected from the above cross. Phenotypes Data Set 1 Data Set 2 Regular Red 26 846 Regular Yellow 15 273 Potato Red 6 287 Potato Yellow 12 94 Total 59 1 500   Convert the expected phenotype ratio into the expected probability for each of the four phenotypes and record them in the table below. (Record your answers as a value between 0 and 1 rounded to two decimal places.) Example of calculating expected probability:To convert from the expected phenotypic ratio to the expected probability…

Genes A, B, and C all assort independently. For the cross AaBbCc X AaBbcc, what fraction of the progeny would be expected to have the ABc phenotype?

In the following cross, imagine that you have a female fly that has two Xs and one Y due to a nondisjunction event in her mother's germ cells. Draw out what the possible gametes are for both the female and the male and also a Punnett square showing the genotypes, phenotypes, and sex of the possible flies as a result of this cross. You do not need to provide the probabilities of each of these. Red-eyed wi C Ở Red-eyed wt XX Y X Y Meiosis

Pure-breeding parental strains of corn with the purple, smooth and yellow, wrinkled phenotypes were crossed to produce an F1 generation, which were then crossed to produce the F2 generation. Complete the diagram showing the results of the cross below. (I included an enlarged picture of the corn as well)

A D. melanogaster male is homozygous for a translocation between chromosomes 2 and 3.  This fly is otherwise wild-type.  This male is crossed to a female fly that is homozygous for normal chromosome order and is homozygous for dumpy wings (dy), a recessive on chromosome 2 and sepia eyes (se), a recessive on  chromosome 3. A male progeny from the above cross is test-crossed to a female fly that is homozygous for dy and se.  What progeny phenotypes and frequencies are expected from this second cross? Answer choices 1/2 wild-type and 1/2 dumpy and sepia 1/4 wild-type, 1/4 dumpy, 1/4 sepia, 1/4 dumpy and sepia All dumpy and sepia                       2/3 wild-type, 1/3 dumpy and sepia 9/16 wild-type, 3/16 dumpy, 3/16 sepia, 1/16 dumpy and sepia