Introduction: The objective of this lab was to study the trait of aldehyde oxidase (AO) in fruit flies. Aldehyde oxidase is responsible for catalyzing the oxidation of many aldehydes. The aldox gene controls the amount of AO activity in Drosophila melanogaster. In the first part of the lab, an enzyme spot test will be performed on two different vials of Drosophila to exhibit the AO activity of both vial 1A and 1B. A positive test for AO test will present a blue color, while a negative test will present no reaction.
Beanium Isotope Lab Introduction: Isotopes are explained as the variations of the number of neutrons that an element may have. Some isotopes are more common than others. This experiment was performed to help visualize the different isotopes of an element and show how some isotopes will appear more often than others. Purpose: To visualize and understand isotopes Materials: Refer to Lab Sheet “Isotope Experiment- Beanium”
Dr. Colleen Winters – BIO 655 Vishall G. Kaistha TITLE: “Recombination-Directed DNA Repair Promote Homologous Stimulating Transcription of Genes That That Preserves Genomic Integrity by MEN1 Is a Melanoma Tumor Suppressor”.
Over these 15 consecutive days our criminal justice class has watched the decomposition of 9 still born piglets, which were laid out in a specific way, with specific wounds. Piglet 10 was put into a box at the beginning of the experiment, and was not revealed until the end of it. Each piglet was in its own condition. Piglet 0 was naked in a box, Piglet 1 was naked on the grass, piglet 2 had 1st, 2nd, and 3rd degree burn all over its body, it too was laying on the grass, piglet 3 was stabbed behind right shoulder, also lying on the grass, piglet 4 had a .380 bullet would on its side, it was laying on the grass, piglet 5 was clothed on the grass, piglet 6 was naked piglet on the black top, piglet 7 was clothed on the black top, piglet 8 was suspended in the air with clothes on, and piglet 9 was suspended in the air naked. At the beginning of the experiment we hypothesized on theories we now know the answer too.
LABORATORY REPORT Activity: Enzyme Activity Name: Natalie Banc Instructor: Elizabeth Kraske Date: 09.26.2016 Predictions 1. Sucrase will have the greatest activity at pH 6 2. Sucrase will have the greatest activity at 50 °C (122 °F) 3.
The data observed and recorded in this lab shows that the concentration of miracle gro’ does affect the growth rate and germination speed of black eyed peas. The data is shown through two graphs and two data tables. The control group in this experiment is the seeds with a 0% concentration of miracle gro’, therefore the seeds with just water. The experimental groups are different concentrations of miracle gro’ including a 10%, 15%, 20%, 25%, and 30% concentration. The variable in this experiment is the amount/concentration of miracle gro’.
Identification of bacteria within Unknown Culture #21 In this experiment, an unknown culture of two different types of bacteria was assigned to each person, a number of tests were performed to isolate and identify these bacterial cells. Based on knowledge from the previous experiments completed in lab, a basic understanding of each type of bacteria was used to create a flow chart that would aid the process of identifying the unknown bacteria within the culture. A gram stain that is performed initially will narrow down the types of tests certain bacteria will and will not respond to. In addition to the gram stain, some of the tests that were used include, a catalase test, an Eosin methylene blue (EMB) agar test, a bile esculin test, and a 6.5% sodium chloride (NaCl) test.
It was used because of its rapid rate of reproduction, The objective of this lab was to determine whether the two crosses conducted fit the expected results of a 9:3:3:1 ratio, which could be done using the Chi Squared Test. The 9:3:3:1 ratio meant that the phenotypes of the F2 offspring would occur in a nine: three: three: one ratio- 9 would be wild-type or normal, 6 of the offspring would have one normal and one mutant characteristic and lastly one of the offspring would be a mutant for both phenotypes. Two distinct crosses were conducted during this experiment: one using red eyed, ebony bodied males and white eyed, brown bodied females as the F1 generation and the other cross using white eyed, brown bodied males and red eyed, ebony bodied females. Our hypothesis, or prediction was that crossing offspring from the F1 generation, or the F2 progeny would fit the expected outcome of a 9:3:3:1 ratio as
Science 1. Free ears in dogs are controlled by dominant allele (F), and attached ears are controlled by the recessive allele (f). In addition, Short dogs is due to a dominant allele(S), and long hair is due to a recessive allele (s). Which of the following is the genotype of the dogs with free ears and short hair? a. ffss b. FfSs c. ffSs d. Ffss 2.
This means that the G0 female population had the genotype w+w+ for red-eyed and ww for the white-eyed fruit flies. On the other hand, male fruit flies only have one X chromosomes with the other sex determining chromosome is Y. Red-eyed males would have the genotypes w+Y and the white-eyed males would have the wY genotype. This means that when looking at the male fruit fly it can be easy to determine the eye color allele with the naked eye. Due to the gene being X-linked the allele frequency for both eye color alleles can be calculated for males, unlike females as a red-eyed female could have either the w+w+ or the w+w genotype ().The Hardy-Weinberg equilibrium states that the genotype and allele frequencies within a population will stay constant from each generation when there is a lack of evolutionary influences, such as genetic drift, mutation, and mate choice
In dogs, rough hair (H) is dominant to smooth hair (h). (3pts) Genotype Phenotype Homozygous dominate Homozygous recessive Heterozygous 4.
Subsequently, the lab group answered: “Would it be possible to have had sexually mature parents with either of the traits expressed in the offspring? Explain (Inheritance of a Single Trait - Lab 9, p. 131).” The group decided that yes, it would be possible for the parents to have been sexually mature and still possess the traits necessary to produce an albino offspring. However, the only way this would have been possible was if the parent’s genotype carried the recessive g gene, but the phenotype of said parent or the actual parent was green. If the parent had been albino, and thus lacked
In this case yellow is dominant and green is recessive. Pure breed make offspring that had green and yellow alleles but only yellow show because it was dominant. For that when F1 generation breed there was change to green alleles to reappear. Alleles are variation of a gene. Also, to represent dominant alleles Mendel use capital letters and recessive alleles are lower case.
During this past week, we covered chapter 2.3, which included topics such as Hardy-Weinberg. Allele frequency, and adaptive evolution. I learned more about microevolution which is a change in allele frequencies in populations over generations on a very small scale. We typically see microevolution during our lifetime, with its main causes being genetic drift, gene flow, and natural selection. Furthermore, we discussed and covered genetic variation among individuals.
This means since my father’s earlobes are attached, differing from my mom’s free hanging ones, one of them has a homozygous recessive genotype. Due to the fact my brother and I both have free hanging earlobes like my mother, the probability is higher that free hanging earlobes are a dominant trait. For example, if my mom’s trait was heterozygous, meaning her phenotype would still be that of a dominant one, the probability my parents have a child with free earlobes is 50% along with the chance of having a child with attached at also 50 %. If my mom carried the homozygous dominant alleles instead of the heterozygous, the probability my parent’s offspring would have free hanging earlobes would be
