ipl-logo

Fruit Fly Lab Report

794 Words4 Pages

The mode of inheritance for the four traits of eye shape, eye color, bristle morphology, and body color were studied in Drosophila melanogaster, or commonly known as the fruit fly. Each gene was analyzed through reciprocal crosses of wild type alleles and mutant alleles in the flies. The gene for body color encodes either a honey yellow or ebony color in the fly. It was determined that the honey, which is the wild type, is the dominant allele and the ebony color is the recessive allele. Eye color, eye shape, and bristle morphology were all determined to be sex linked on the X chromosome as they are more prevalent in the males. If these genes were found on the Y chromosome, there would be no females found with mutations in the eye shape, eye …show more content…

As seen in Figure 1, the linkage map displays the map units and the order of the eye shape, eye color, and bristle type genes. In order to determine the distance between the genes, map units must be calculated, which can be found in Table 5. The calculations were performed by adding together the recombinants of the parents and dividing by the total number of flies counted and multiplying that by 100. This was done for each of the genes in order to identify which of the three genes was in the middle. Determination of the middle gene allowed for the relative positions of the other two genes to be calculated. Bristle morphology was determined to be in the center of the other two as it is the gene that switches in the recombinants. Bar eyes was calculated to be the first gene on the X chromosome 47.0 map units away from the middle gene bristle morphology. Eye color was found to be 17.7 map units from the middle bristle morphology gene, and 64.7 map units away from the bar eye gene, making it the last of the three genes on the X chromosome. The body color gene was already found to be on a separate chromosome. Therefore, it is not linked to eye color, eye shape, or bristle morphology that were found to be on the X …show more content…

The cross of BWSE X ++++ were found to be BWS+. Female F1 were found to be B+++ for the cross of ++++ X BSWE and for the cross of BWSE X ++++. The absence of an ebony body color in both the male and female F1s corresponds with the recessiveness of this allele. Bar eye can be observed in the females due to the distance between the linked genes. There is a higher chance of the eye shape allele of the chromosome to switch as this section is 47.0 map units away from the bristle morphology and even further from the eye color. The body color gene in the F2 data was analyzed using a chi square test to see if the it fit the expected 3:1 Mendelian ratio. There were 168 honey body colored flies observed in the F2 data, and 47 ebony body colored flies. These two numbers were compared to expected values for a 3:1 ratio in order to obtain the X2 value of 1.213. The degree of freedom was determined to be 1 for the data, which is necessary for the interpretation of the X2 value. Using the degree of freedom and the X2 value, a corresponding p-value of 0.2710 was determined. This p-value is above the Alpha value of 0.05 indicating that the data does not differ significantly from the 3:1 ratio. In addition this p value greater than 0.05 implies that any observed deviation is obtained by chance alone, specifically between 20% and 30% of the time. Therefore, the null hypothesis of the F2 data fitting the expected Mendelian 3:1 ratio

Show More
Related
852 Words | 4 Pages

Later it was discovered that it was the result of an extra copy of chromosome 21. The nondisjunction that results in an extra copy of chromosome 21 occurs during anaphase I in meiosis I. The genetic mutation is trisomy 21 (3 copies of chromosome 21). The characteristic phenotypic occurrences that are distinct to the disorder: poor muscle tone, stout neck, flat face, small head, mouth, and ears, eyes slanting upwardly, Brushfield spots, and stout fingers and

Read More
532 Words | 3 Pages

There was no indication of phenotypic differences between male and female mutant flies ,illustrating that the mutation was not sex linked because both mutant sexes possessed the same eye mutation. Male flies in both wild type and mutant flies consisted of the sex combs in the front of the legs, and dark coloration located at the end of the abdomen. Both classes of female flies possessed a tiger- striped abdomen and a tag-like tip at the end of the abdomen. After confirmation of our eye mutation we can assume that our mutation can be due to either genes rosy, garnet, brown, purple, and sepia because these are genes that affect the pigmentation color of the Drosophila melanogaster. Our mutation may be caused by the transportation of the enzyme xanthine dehydrogenase to the rosy gene which causes a deficiency in the pigmentation

Read More
1080 Words | 5 Pages

Materials: 1. Genetic code chart 2. The Making of the Fittest: Natural Selection and Adaption Student Handout

Read More
947 Words | 4 Pages

This experiment was conducted to determine whether or not Callosobruchus maculatus, or bean beetles, had a bean color preference for oviposition choice. Oviposition is the process of a female insect laying her eyes on plant parts and other materials, which can be influenced by many factors. The bean beetle eggs are opaque and clear, which allowed us to test the hypothesis that C. maculatus prefer the darker red Adzuki beans over the white Black-eyed peas for oviposition choice. Two different colored bean types were used, the red Adzuki beans and the white Black-eyed peas. We placed three female and two male bean beetles in each petri dish, with 55 of each bean type randomly placed in the dish, for a week.

Read More
708 Words | 3 Pages

In this experiment Drosophila melanogaster (fruit fly) were used. To begin with the lab, four virgin females with mutant genotype for yellow body color, crossveinless, and forked bodies were mated with genotype (y cv v f / y cv v f). Then these females were crossed with six males with all wild-type genotypes, which can be abbreviated as (+ + + / Y). During the first week of experiment the flies were anesthetized and viewed under dissecting microscope. Then the flies with above characteristics were chosen.

Read More
471 Words | 2 Pages

The expected results of these crosses if the genes assort independently, are that all the offspring from the P generation would have wild-type allele with red eyes, long wings and dark bodies, the flies would be heterozygous for each allele. If the genes were sex-linked and have females in P-generation, the traits in F1 generation in males would be wild-type. In the reciprocal cross, males would exhibit phenotypes of female if it is mutant phenotypes because only female would inherited her X chromosome to the male in new generation. The adults from the P generation were removed before the F1 generation developed from

Read More
656 Words | 3 Pages

These tubes were then labeled and then placed into an incubator where they were able to mate. After a week the flies were anesthetized once more and then the F1 generation was removed and all the pupa were placed in tubes. After another week past the F2 flies were anesthetized again and documented based on whether they showed the sepia gene. This time all the flies that had sepia eyes were removed from the tube. Next, 10 red eyed F2 male and female flies were placed in a tube to mate and help form the next generation.

Read More
942 Words | 4 Pages

The purpose of this experiment was to determine how the color genotype is carried on the chromosome in the Drosophila melanogaster (fruit fly). The results of the first generation (F1) did not show what the hypothesis predicted. Only white eyes were supposed to be present, yet the F1 generation had a clear split between the generations with red eyed females and white eyed males (Table 1). There was no other eye color present for either sex. The F2 generation resulted in both male and female flies having either red, white, or sepia eyes (Fig. 1).

Read More
473 Words | 2 Pages

Drosophila is a fly species that is used in genetics. Since Drosophila goes through life cycles extremely quickly, it is one of the best specimens to use in a multiple generation experiment. The purpose of this experiment is to follow the genetic phenotype of the offspring and determine the genotype, which will most likely predominately be the wild type phenotype with a few recessive traits. To conduct the experiment, we first cross the parents, wild type with sepia eyes and vestigial wings or black body with vestigial wings. The offspring from that cross is expected to be 100% wild type and the following cross between the F1 generation is expected to be 9:3:3:1.

Read More
1720 Words | 7 Pages

(1) Ecologists can distinguish between them because the stable fly has

Read More
189 Words | 1 Pages

The green allele represents p, and q is 0.4 The expression goes like p + q= 1 We subtract 1 by 0.4 (1-0.4) and we get the frequency of F which is 0.6 C. 0.48 Using the Hardy-Weinberg equation to find the heterozygotes, is p (in this case the frequency of F) and q (which in this case is the frequency of f) multiplied by 2 p is 0.6 q is 0.4 (0.6)(0.4) x 2 and we get 0.48 D. 0.16 The question answered it by saying that 16 have gray fur, and those with gray fur are the ones with the homozygous recessive alleles and since 16/100 that means it's 0.16 E. 0.36

Read More
697 Words | 3 Pages

White colonizers moving westward posed a huge threat to Native American tribes in the late 18th and early 19th centuries. These tribes were threatened with the loss of their homes, ways of life, and families when white people came to forcefully eject them from land that was rightfully theirs. Tecumseh, a Shawnee leader, recognized that the only way to defeat the violent white men was to unite his tribe with other tribes faced with the same problems. Tecumseh met with the neighboring tribe, the Osages, to deliver a speech calling for the unification of the tribes. Through his use of figurative language, diction, and allusion, Tecumseh attempts to unite the Shawnee with the Osages against their common enemy, the white man.

Read More
1617 Words | 7 Pages

The purpose of this investigation is to determine the inheritance pattern of the albino trait and whether the albino gene in corn seedlings is dominant or recessive. One group of lab partners in a lab class examine a tray of corn seedlings showing both green and albino plants. The students record their findings and compare their tray of corn seedlings’ results to five other lab groups in the class. The lab students used this information to test all of the possible gene combinations the parents that produced partial green and albino offspring may have had, using Punnett squares. Based on their findings, the groups determine the hypothetical genotype and phenotype ratios of the corn seedlings and use this to form a hypothesis.

Read More
820 Words | 4 Pages

For the most part, the only relation race has to genetics is the indication of an individual’s ancestor and their skins adaptation to a geographic

Read More
707 Words | 3 Pages

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

Read More

More about Fruit Fly Lab Report

Related Topics

Open Document