To manipulate the flies, they are anesthetized using CO2 that is administered through the FlowBuddyTM device with a 5 L/min flow rate. Once immobile, the flies are examined and scored under a 10.5-45X magnification stereomicroscope. The flies are returned to the incubation chamber set at 25ºC to facilitate optimal growth conditions. The first discriminant cross, DC1, is made by crossing mutant virgin females with Curly/Plum; Dichaete/Stubble marker males. Depending on the genetic outcome of DC1, the second discriminant cross, DC2, is made by crossing female wild-type virgins with F1 males if the unknown mutation is dominant, or by crossing F1 males with mutant females if the unknown mutation is recessive.
a) Bb x Bb c) BB x BB b) BB x bb d) Bb x bb 5) When a certain pure strain of fruit fly is cultured at a temperature of 16˚C, all of the flies will develop straight wings. The offspring of these flies will develop curly wings when they are raised at 25˚C. This pattern of development indicates that a) acquired characteristics can be passed from generation to generation b) curly-winged flies will have straight- winged offspring regardless of the environment c) the environment can affect the expression of certain genes d) straight-winged flies will have curly- winged offspring regardless of the environment The changes in foot structure in a bird population on over many generations are shown in the diagram below. 6) These changes can best be explained by the concept of a) evolution b) extinction c) stable gene frequencies d) use and disuse 7) In 1889, August Weismann, a German biologist, conducted an experiment attempting to produce mice without tails. He cut the tails off adult mice and then permitted them to mate.
bb x bb e. BB x bb 29. To determine whether an unknown black guinea pig is pure or hybrid black, it should be crossed with a. a white b. a hybrid black c. a hybrid white d. a pure black e. another unknown 30. Organisms that have two identical alleles for a particular trait are said to be a. hybrid b. homozygous c. heterozygous d. dominant 31. A pea plant heterozygous for height and seed color (TtYy) is crossed with a pea plant heterozygous for height but homozygous recessive for seed color (Ttyy). If 80 offspring are produced, how many are expected to be tall and have yellow seeds?
b) In the data table, record the tally for each phenotype in the row marked “Simulated Number.” c) In the row marked “Simulated Percentages,” convert the simulated numbers (out of 10) to simulated percentages (out of 100%). d) Compare the predicted values to the simulated values. How does the data confirm that black fur and black eyes are indeed dominant alleles? Step 3: Predict the Offspring of a Dihybrid Cross (BbEe × BbEe) Repeat step 1 using two mice from the F1 generation that are heterozygous for both fur color and eye color (BbEe). Step 4: Simulate the Cross in Step 3 Repeat step 2, except for two mice that are heterozygous for fur color and eye color (BbEe).
The long, thin ascus causes both meiotic spindles and the mitotic spindles to be aligned during spore formation. As a result, the positioning of the ascospores directly reflects the arrangement of homologous chromosome pairs during metaphase I of meiosis (Glase, 1995). This allows us to detect which alleles were exchanged creating crossovers. Sexual reproduction in the S. fimicola is important because an it allows for a significant source of genetic variability, and allows the fungus to adapt to new environments (Fincham, 1971). This genetic variability is obtained when crossovers occur between
B B Bb | Bb | Bb | Bb | b b All (Bb) heterozygous genotypes are expressing the phenotype of black eyes. If we do another cross for F2 generation we will find 2 heterozygous black eyes, 1 homozygous black eye and 1 homozygous brown eye among the offspring. The punnet square is as follows: B b BB | Bb | Bb | bb | B b The genotypic ratio is – 1(BB):2(Bb):1(bb). The phenotypic ratio is – 3:1. DROSOPHILIA MELANOGASTER AS A MODEL FOR STUDYING GENETICS Drosophila melanogaster also known as the fruit fly is used commonly in the study of genetics.
Genevieve 171L Honors 11/6/08 Characteristics of the Aldox Gene Introduction This experiment was conducted by using aldehyde oxidase (AO) to genetically characterize the aldox gene. It was hypothesized that the aldox gene is recessive and sex linked. Fruit flies (Drosophilia melanogaster) are genetically well known eukaryotic organisms that were used to investigate the inheritance of the aldox gene. To determine if AO is present, an enzyme spot test was used. The yellow color indicated that AO was present and active while blue color indicated AO is not present.
Fern Spore Germination and the Effects of Ammonium Nitrate Name Lab Partner: U-# BSC: 2011L Section: 908 April 10, 2014 Abstract: The effect of 1% ammonium nitrate on Ceratopteris richardii spore germination was tested in this experiment. Past experiments have been done on Botrychium dissectum and they found that when the spores were grown in the presence of ammonium, they had the best growth rates. Also when they had ammonium nitrate present, even more spores germinated compared to when they only used nitrate by itself. They did several tests using nitrate, ammonium, and then a combination of ammonium nitrate. Basing our hypothesis off of this past experiment, I hypothesized that in the presence of ammonium nitrate we should see an increased germination rate of our spores from the Ceratopteris richardii.
I observed that the females in the F1 generation all had wild type phenotypes and the males in the ratio all had tan body phenotypes. There was a ratio of about 1 to 1 of female wild type flies to male flies with tan bodies. ENTER DATA: (cross between two F1 flies from the above cross—this will produce the F2 offspring) Phenotype Number Proportion Ratio Female: + 248 0.2518 1.046 Male: + 237 0.2406 1.000 Female: T 261 0.2650 1.101 Male: T 239 0.2426 1.008 Total 985 2. Question: Based on what you know about Mendelian genetics, did the F2 generation demonstrate the phenotypic ratio that you expected? If not, what phenotypic ratio was obtained with this cross, and why?
i. Earthworm tasting ability j. One thumb k. Both of the above l. None of the above 4. In the homologous pair of chromosomes above, what traits is the organism heterozygous for? m. Saliva stickiness and number of thumbs n. Whether it has polka dots and its earthworm tasting ability 5. Which of the following accurately describes the phenotype of the organism?