Cytochome-C Comparison Lab

Cytochome-C Comparison Lab 



PURPOSE: To compare the relatedness between organisms by examining the amino acid sequence in the protein, Cytochrome C.

Cytochome-C
Small protein from eucariotic cell,
Associated will in member of the mitocondray,
 Is molecules oxidizable ,
 Hemoproteinthe
Funcion part of electon transporchain produce energi (ATP)

This test will see the difference in cytochrome-C have the following animals

Horse -Donki- - MAMMALS
Chiken-Penguin- BIRDS
Snake-REPTILE
Silkworm moth-INSECT
Yeast-FUNGI
Wheat-PLANTS

METHOD 

Compare the amino acid sequence of Cytochrome-C in various organisms.
. Count and record the total number of differences.
Share your data with the rest of the class and complete
Make a branching tree, or cladogram with results


TABLE

Differences between amino acid sequences in Cytochrome C protein for nine species.

	Horse	Donkey	Whale	Chiken	Penguin	Snake	Moth	Yeast	Wheat
Horse	0
Donkey	0	0
Whale	5	4	0
Chiken	11	10	9	0
Penguin	13	12	10	3	0
Snake	21	29	18	18	19	0
Moth	24	23	22	23	23	26	0
Yeast	40	39	39	40	39	40	46	0
Wheat	38	37	36	39	39	37	40	43	0

Cladogram

Image the blog ines: llimona http://ilsciencelab.blogspot.com.

QUESTIONS

1-) How many Cytochrome C amino acid sequence differences are there between chickens and turkeys? 0
2-)Make a branching tree, or cladogram for chickens, penguins, and turkeys.

3-) Predict the number of Cytochrome C amino acid sequence differences you would expect to see between
a). horse and zebra? 1-2 aprox
b). donkey and zebra?1-2 aprox



4-) What other information did you use to make this prediction?

	Horse	Penguin	Snake	Tuna	Moth	Yeast	Wheat
Tomato hornworm moth	26	18	24	28	5	42	38

5-) List three other things used to determine how organisms are related to each other? If they can reproduce and the offspring is fertile compering organs proce anatomic

6-) Explain why more closely related organisms have more similar Cytochrome C ? Are evolutionarily and not as separate DNA canviado not therefore less time with less canvios occur in the evolution of the species

7-) Other data, including other genes, suggests that fungi are more closely related to animals than plants.What are some reasons that the Cytochrome C data suggests that fungi, plants, and animals are equally distantly related? Based on a number of mutations as over forty are no longer compatiobles among them, have undergone many mutations

p:10 protein denaturation 1

DENATURATION:


Denaturation: Is a process in which proteins or nucleoids acids lose the quaternay tertiary and secondary structure that is present in their native state.Denaturation is the result of the application of some external stress(heat and pH change ) or compounds such as a strong acid or base a concentrated inorganic salt or organic solvent.

Objectives

1- Study the relation between the structure and the funcion of proteins.
2 - Understand how temperature,pH and salinity affect to the protein structure.

MATERIALS:
 2x250 ml beaker
4 test tube
Test tube rack
10ml pipet
Knife
Glass marking pen

Potato
Distilled water
Hydrogen Peroxide
NaCL
HCL

PROCEDURE


In this experiment we are going to test catalase activity in different environmet situation.We are going to mesure the rate of enzyme activity under various conditions ,such as different pH values and temperatures.We will measure catalase activity by observing the oxygen gas bubbles when H2O2 is destroyed .If lost of bubbles are produced, it means the reaction is happening quickly and the catalase enzyme is very active.

1 Prepare 30ml of H2O2 10% in a beaker
2 Prepare 30ml of  HCL10% in a beaker
3 Prepare 30ml of NaCL 50% in a beaker
4 Peel a fresh potato tuber and cut the tissue in five cubes of 1cm .Weigh them and equal the mass
5 Label 5 test tubes (1,2,3,4,5)
6 Immerse 10 minutes your piece of potato inside HCL beaker
7 Immerse 10 minutes another piece of potato inside NaOH beaker
8 Boil another piece of potato
9 With a mortar ,mash up the third piece of potato
10 Prepare 5 test tubes as indicated below:

TUBE	TREATMENT	Resuls potato+H2O2	Cm long potato+H2O2
1	Raw potato	treatment had not	6cm
2	Boiled potato	no bubbles have appeared	3,5cm
3	Potato with HCL treatment	had to be almost no activity	3,8cm
4	Potato with NaCL treatment	been oxidized	3,4cm
5	Mashed up potato	to have more surface contact has more reaction	6,8 cm

11 Add 5ml H2O2 10% in each test tube
12 With a glass-marking pen mark the height of the bubbles.Measure it with a ruler
13 Compare the results of the 5 test tubes.

Complete the table below with the important parts of this experiments:

Parts:	In this experiment this was....
Independent variable	Treatment of each potato,pH,saliniti, temperature
Dependent variable	Height of the bubbles
Experimental group	2,3,4,5
Control groups	1
Constants	5ml H2O2, time , 1g of potato

Represent your results in a chart: Treatemer vs bubble's height:

QUESTIONS

1 How did the temperature of the potato affect the activity of catalase?
2 How did the change of pH of the potato affect the activity of catalase?
3 In which potato treatment was catalase the most active? Why do you think this was?
4 An experiment was performed to test the effect of temperature and pH on the activity of    Enzyme X.The following data was collected during the experiment:

a) What is the optium pH of enzyme X?
b)What is optium temperature of enzyme X?
c) Why do you think enzyme X has low activity at a pH of 10?
d)Enzyme X permoms critical life functions.Use the data adove to explain why a fever of 40 degrees may be dangerous.