The balancer chromosome includes GFP and a mutation in gene X that causes flies to be sterile in the homozygous state. The recessive mutation that you are studying is in gene Y, and causes flies to develop two tail ends in the homozygous state. This mutation is lethal in the homozygous state. For each of the following phenotypes in the balanced mutant line, what alleles are present and in how many copies? a.) Glows green, fertile (Select) b.) Glows green, sterile ISelect) c.) does not glow green, two tails ( Select] If you crossed a fly from your balanced line with the following flies, what proportion of the offspring would glow green and be fertile? d.) wild-type ISelect ) e.) fly with balancer chromosome from parental generation [ Select ]

Consider a maize plant: Genotype C/cm ; Ac/Ac+ where cm is an unstable colorless allele caused by Ds insertion.  What phenotypic ratios would be produced and in what proportions when this plant is crossed with a mutant c/c Ac+/Ac+?  Assume that the Ac and c loci are unlinked, that the chromosome-breakage frequency is negligible, and the C allele encodes pigment production.

In c. elegans, genetics model organism, movement problems (unc) and small body size (sma) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (unc+ and sma+). A worm homozygous for movement problems and small body is crossed with a worm homozygous for the wild-type traits. The F1 have normal movement and normal body size. The F1 are then crossed with worms that have movement problems and small body size in a testcross. The progeny of this testcross is: Normal movement, normal body size            210 Movement problems, normal body size          9 Normal movement, small body size                 11 Movement problems, small body size           193   a)From the test cross results, can you tell if the two genes are on the same chromosome or not? Explain your reasoning.   b)What phenotypic proportions would be expected if the genes for round eyes and white body were located on different chromosomes?  (please explain hot to get to these conclusions)

A complementation analysis was performed using Drosophila which normally have black wings. Six mutants for clear wings were discovered. The mutants were bred in a complementation test to produce the following: Mutant 2 4 # 1 1 1 1 2 1 1 1 1 1 1 4 1 1 Where a "0" (zero) means clear wing offspring and a "1" (one) means wildtype (black wing offspring). How many genes can you identify that affect the production of wing colour in Drosophila? O a. 3 O b.4 О с. 5 O d. 6 O e. none of the above 3. 5

A homozygous fly with two recessive mutations causing purple body and short wings (ppss) is mated with a fly heterozygous for both mutations (p+p and s+s): "+" represents the wild-type allele. They have the following offspring:   26   Wild type (normal body, normal wings) 345   Purple body, normal wings 376   Normal body, short wings   13   Purple body, short wings What is the allele arrangement (haplotype) for the heterozygous parental fly? ("p" designates the mutant allele; "p+" designates the wild-type allele.)

There are two genetic disorders that result from mutation in imprinted genes: Prader-Willi syndrome, Angelman syndrome.  Angelman syndrome results from deletion of UBE3A, which is a gene imprinted such that only the maternal copy is expressed. In the pedigree above, individual I-1 is heterozygous for a deletion of UBE3A and does not have Angelman syndrome.  Individual I-2 is homozygous wild type for UBE3A.  Which individuals in the pedigree are at risk for exhibiting Angelman syndrome, if any?  (Who could potentially have the syndrome, based on what alleles it is possible for them to inherit and express?)   Question 8 options:   Only I-1 could have been at risk.  If he does not have the syndrome, no one in the pedigree could.   Only III-1 is at risk   I-1, II-2, and III-1 are all at risk   Only II-2 is at risk   No one in the pedigree is at risk   Both II-2 and III-1 are at…

The data set attached presents the results of a testcross using female flies heterozygous for three traits and male flies, which are homozygous recessive. For simplicity, mutant alleles are shown with letters a, b, and c and wildtype alleles are indicated by a “+” symbol. For this part of the report do the following in order: a) Determine the gene order (which gene is in the middle?)d) Construct a genetic map for the three genes, including the map distances between them. Clearly indicate the logic you followed and show all your calculations. Include the full distance calculations for the two most distanced genes (do not just add the other 2 distances).  Ensure the work is neat and clear and does not contain spelling or grammatical errors so that it is understandable. Make sure to double check the solution provided.

In a wild-type fungus, protein E (encoded by the haplosufficient gene E) normally dimerizes to catalyzes a biochemical reaction necessary for the production of a dark pigment. Ed represents a mutant, dominant negative allele of gene E. What is the predicted phenotype of a fungus cell of genotype E*/Ed, and why? O wild type (normal production of the dark pigment), as E is haplosufficient mutant (no pigment production), as no dimers will form in the heterozygous mutant (no pigment production), as the mutant allele Eg is dominant O wild type (normal production of the dark pigment), as dimers of wild-type and mutant protein E will be formed in the heterozygous

Lid (Little imaginal discs) is a gene located on chromosome 2L and FRT40A refers to an FRT sequence inserted at the 40A (cytological location). Tubulin (Tub) refers to a promoter that can drive target gene expression in all the cell types. Gal80 is an inhibitor of Gal4 that blocks Gal4’s ability to activate transcription. Determine whether the following statements are true or false for flies with the following genotype: y w HsFLP; lid, FRT40A / P{Tub-Gal80}, FRT40A; P{Tub-Gal4} P{UAS-GFP} / + A. The fly will express Gal4, which will activate GFP in all the somatic cells without heat shock. B. After heat shock, FLP recombinase will be induced, which will induce mitotic recombination at FRT40A. However, lid mutant clone will be not induced because it is located closer to the centromere than FRT40A. C. After heat shock, FLP recombinase will be induced, which will induce mitotic recombination at FRT40A, resulting in the generation of homozygous lid mutant clones and twin spots that are…