Genetics Lab Exam I

Front 1

Why is Drosophila useful as a model organism?

Back 1

-Simple and cheap breeding
-Short life cycle
-High number of offspring
-Polytene chromosomes

Front 2

Polytene Chromosomes

Back 2

-large chromosomes in salivary glands of Drosophila
-many cell cycles w/o division
-characteristic banding pattern
-can see it during interphase

Front 3

Life Cycle of Drosophila

Back 3

-Fertilization internal
-Female stores transferred sperm for several days
-Male sexually mature in ~6hrs.
-Female sexually mature in ~8 hrs.
-Eggs deposited on surface of food source
-1 day egg hatch and 1st larval stage
-Arthropod - invertebrate having jointed limbs and a segmented body with an exoskeleton made of chitin
-As grows sheds exoskeleton

Review Slide
egg --> 3 larval instars --> pupa --> imago --> adult fly

Front 4

Calculate % Recombinance between 2 genes

Back 4

Example: Eye Color vs. Bristle Shape
White Straight --> 19+26+10+2
Red Curly --> 16+13+11+7

--> (104/523) x 100 = 19.9% --> linked

*50% is the limit of the test

Front 5

Calculate Interference

Back 5

I = 1 - Co

ex. --> I = 1 - .77 = .23 --> 23% of the time expected DCO's are not seen

Front 6

Coefficient of Coincidence

Back 6

=(observed DCO/expected DCO)

expected DCO --> (freq. cross over between eye color & bristle shape)(freq. cross over between bristle shape & eye color)(total flies)

ex --> observed/expected= 30/39 = .77 --> 77% of the time we see DCO when expected

Front 7

How can you tell which loci are autosomal and which are X-linked

Back 7

-for X-linked loci, F1 males and F1 females have different phenotypes --> F1 males have same phenotype as mutant mother because he inherits his one X chromosome from her

-For autosomal loci, F1 males & females are both heterozygous & have the same phenotype

Front 8

Polytene Chromosome Characteristics

Back 8

-Polytene chromosomes have undergone several rounds of DNA replication without cell division
-Polytene chromosomes are very large chromosomes that can be seen during interphase when cells are metabolically active
-Polytene chromosomes are paired with their homolog in somatic cells (unusual) which makes them appear even larger
-The banding pattern is specific to each chromosome which makes it easy to observe large chromosomal aberrations such as deletions, insertions, translocations and inversions
-Each band spans about 30,000 bp – enough DNA to encode several genes

Front 9

Why did we use Drosophila virilis for the polytene chromosome lab?

Back 9

Drosophila virilis is a larger fly & therefore has larger salivary glands that are easier to isolate

Front 10

Hyphae

Back 10

tiny filaments in fungi

Front 11

Mycelium

Back 11

interwoven mat of hyphae

Front 12

Ascospores

Back 12

form through meiosis in a sac called an ascus

Front 13

Why is studying a haploid organism useful?

Back 13

can see recessive mutations immediately

Front 14

Plasmogamy

Back 14

the union of cytoplasm of two gamete cells

Front 15

Karyogamy

Back 15

the union of the nuclei

Front 16

Dikaryon

Back 16

the state in certain fungi in which each compartment of a hypha contains two nuclei, each derived from a different parent

Front 17

Ascomyocytes

Back 17

-Karyogamy occurs within an ascus and the diploid nucleus divides by meiosis
-one mitotic division follows meiosis to produce 8 ascospores in each ascus

Front 18

Spores

Back 18

-single cell
-haploid
-resistant to harsh conditions

Front 19

Antheridium

Back 19

-acts as male
-donates nuclei to ascogonium

Front 20

Ascogonium

Back 20

acts as female

Front 21

Asexual Reproduction

Back 21

creates genetically identical spores

Front 22

When counting asci in Sordaria, which arrangements go together?

Back 22

-4:4 arrangement
-2:2:2:2 + 2:4:2 arrangements

Front 23

Calculate freq. of recombinant asci

Back 23

ex. --> tan = (2:2:2:2 + 2:4:2)/total asci = 312/573 = .54 --> percentages of recombinant spores = .54/2 = 27%

Front 24

Polytene Chromosome Staining Procedure

Back 24

-place 3rd instar larva on slide with saline solution
-hold larva down and pull needle in head away from body to remove salivary glands
-allow glands to soak in saline for 10 minutes (then blot it up)
-place aceto-orcein stain on glands and let site for 15 mins
-red nuclei appear
-blot up old and place new drop of stain
-squash to break open nuclei
-observe chromosomes

Front 25

Homologous Chromosomes Pair in what stage for polytene chromosomes

Back 25

prophase I of meiosis I

Front 26

Complementation Test

Back 26

-can determine if these mutations are alleles in the same gene or mutations in different genes

Front 27

Compensation Test

Back 27

-can determine which steps (enzymes) cn and v are related to in the biochemical pathway

Front 28

Importance of brown mutation in Drosophila

Back 28

The brown mutation is involved in making red pigment. Being homozygous for the brown mutation means there is no red pigment made. The brown pigment is easier to see in the absence of red pigment. With the brown mutation, one needs to distinguish between brown, peach, and white eyes. Without the brown mutation, one needs to distinguish between deep red & bright red eyes. Without the brown mutation, one might not be able to distinguish between the cinnabar & vermilion phenotypes.

Front 29

Are the mutations in the same or different gene pathways?

Back 29

-the phenotype of the cn/bw x v/bw F1 flies is brown eyes
-This means the mutations are in different genes in the pathway
-Since the F1 flies habe one working copy of the gene to make E1 and one working copy of the gene to make E2, the F1 flies show the wild type phenotype of brown eyes indicating that the cn and v mutations are in different gene pathways