• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Fruit Fly Lab

Extracts from this document...


Alice Wang Genetics of Organisms Question: Is an unknown Drosophila mutation dominant or recessive, and is it X-linked or autosomal? Hypothesis: The offspring of a pair of reciprocal Drosophila crosses is determined by which the mutation is dominant or recessive and whether it is x-linked or autosomal. Materials: * Drosophila cultures * Dry active yeast * Fly nap wands * Ether * Fruit fly media * Small paint brushes * Dissecting microscopes * Glass vials * Foam stoppers Outline: 1. First week: Immobilize and remove the adult flies of the original vial. Observe them carefully under the dissecting microscope. Separate the males from the females and look for the mutation(s). Note whether the mutation(s) is/are associated with the males or females. 2. Place the parents in the morgue( jar containing alcohol). Label the vial containing the eggs or larvae with the symbols for the mating. Also label the vial with your name and date. Place the vial in a warm location. 3. Second week: Begin by observing the F1 flies. Immobilize and examine all the flies. Record their sex and characteristics. Consider the conclusion s that can be drawn from these data. Place five or six pairs of F1 flies in a fresh culture bottle and the rest of the flies in the morgue. 4. Third week: Remove the F1 flies from the vials and place them into the morgue. ...read more.


apterous 8 8 Red x wild 46 29 Sepia x wild 4 9 Cross 4 Results: F2 Generation Phenotype # males # females Red x apterous 2 5 Red x wild 2 16 Sepia x wild 2 1 Cross of Apterous from F2 Generation Cross # Generation Phenotypes male Phenotypes female # males # females Offspring (F3) 5 F2 Red x Apterous, Sepia x Apterous Red x Apterous. Sepia x Apterous 4 1 4 2 sepia x apterous, red x apterous, sepia x wild, red x wild Cross 5 Results: F3 Generation Phenotype # males # females Sepia x wild 3 2 Red x apterous 6 7 Sepia x apterous 4 5 Red x wild 8 11 The data that I have collected is not very reliable. My results indicate that apterous is a recessive trait while the wild type is dominant according to offspring. However, my group made a huge mistake in the F1 generation. For cross three and cross four of the F1 generation, the offspring phenotype should not have contained any wild type fruit flies because apterous winged is a recessive phenotype. Instead of crossing apterous fruit flies, my group mistakenly crossed a different vial of flies. Instead of waiting for the original vial of (parent) fruit flies to lay offspring for the F1 generation, my group crossed the parent fruit flies. ...read more.


Thus, incorrectly identifying the characteristics of the flies could have also greatly affected the results received. Also, instead of waiting for the original vial of (parent) fruit flies to lay offspring for the F1 generation, my group crossed the parent fruit flies. Hence, improper calculation of numbers could have also caused inaccurate results. Finally, some flies could have gotten stuck in the medium and could have been identified. Improvements can be made if the investigation crossed more fruit flies, thus giving a more veritable number of fruit flies for each phenotype and hence a more reliable data. Finally, crossing the F1 generation instead of the parent generation would be more veritable for the data collected. Apterous x sepia X Apterous x wild (ww x rr) x (Ww x Rr) wr wr wr wr wr WR WwRr WwRr WwRr WwRr WwRr Wr Wwrr Wwrr Wwrr Wwrr Wwrr wR wwRr wwRr wwRr wwRr wwRr wr wwrr wwrr wwrr wwrr wwrr Red x Apterous = 7/28 Wild x Wild = 18/28 Sepia x wild = (3/18) The Chi square (?2) test is used for discontinuous values. ?2 = ? ((O-E)2/E) Where E = expected values O = observed values ?= sum of Observed Expected (O - E) (O - E)2 (O - E)2/ E Wild type (18/28) (14/28) 2 Wild type (18/28) (14/28) 2 Table to show Chi square analysis and probability of F1 generation for monohybrid cross Degrees of freedom = number genotypes - 1 2 genotypes - 1 = 1 degree of freedom Critical value = 3.84 ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Misc section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Misc essays

  1. Comparing and Contrasting Envy and Deception in Shakespeares Much Ado About Nothing and Othello

    Indeed Beatrice claims to be against the idea of getting married as she says, "What should I do with him- dress him in my apparel and make him my waiting gentlewoman? He that hath a beard is more than a youth, and he that hath no beard is less than

  2. Proyecto Personal - Anorexia y Bulimia

    Por lo tanto, ten�a bastante informaci�n de diversas fuentes como Internet, libros, entre otras, y tuve que leer detenidamente tratando de captar el concepto central de cada aspecto, para luego redactar mis propias definiciones que incluir�a posteriormente en mi producto.

  1. SWA lord of the flies

    However, I know there is nothing we can do to kill this beast that now lives inside each one of us. It possesses an incredible and unpredictable evil power. It is not the creepy creature that once we saw on the mountain anymore; he has transformed into an invisible creature that cannot be seen but felt.

  2. A comparison and analysis of the speed and efficiency between Microsofts Windows XP and ...

    After BootROM, 'boot.efi' is loaded which loads the operating system. This can be seen as the 'booting' screen. 7 The number of overall processes during startup in the Mac are much less compared to Windows XP. Hence, it can be predicted that this would have an effect on the speed of the respective operating systems accordingly.

  1. Effect of Light intensity on the Anaerobic Respiration of Pygmy Chain Sword

    By taking in carbon dioxide and giving off oxygen as a waste gas, plants which photosynthesize help to replenish the oxygen levels providing organisms around the world with the oxygen that they need to live. Many factors affect the rate of photosynthesis in plants, such as light, pH, and temperature,

  2. Name of the Rose Analysis Questions

    the scene saying, "The abbot ordered the corpse to be extracted from the ghastly liquid" (104). Referring to the blood as a "ghastly liquid" supports the biblical allusion. The second allusion is one referring to a previous phrase made by William concerning the "hoof prints in the snow" (26)

  1. ESS Assess lab

    At habitat adjacent to the pond (mid altitude) : Moisture content = Initial wt. - Dry wt. = 10g - 7.548g = 2.452g of moisture/10g of soil Nitrate Content = 5 ppm ?? Phosphate content = 35 ppm pH = 7.37 Conductivity = 0.001 Soil Analysis : Sr.

  2. GCSE Welsh Baccalaureate Completed Diary Pages

    However, after watching the video, I will take particular attention when walking through towns. I feel more equipped now to recognise the value of being homeless, and no longer feel an urge to turn a blind eye to the homeless.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work