🏷️ LIMITED TIME OFFER: GET 20% OFF GRADE+ YEARLY SUBSCRIPTION →
Log in
HomeStudy Guides420,000CA160,000

Biology 2581B Study Guide - Erwin Chargaff, Oswald Avery, Martha Chase

43 views3 pages
coralimpala296
3 Mar 2014
School
Western University
Department
Biology
Course
Biology 2581B
Professor
Jim Karagiannis

For unlimited access to Study Guides, a Grade+ subscription is required.

Document Summary

Consider a simple game, whereby we have an unknown (x) to be a number between 1 and: in this way, data can only be generated by asking is the number higher or lower than x. In the first round we find that the number is higher than 10, this yields information as the number now has to be in between 11 and 100. In the second round, we find that the number is higher than 0, which yields no valid information as we already know that it is greater than 10 (remember that data does not necessarily equal information). In the third round we find that the number is higher than 5, which like round 2, yields no valid information. In round four we find that the number is higher than 50, which means that the number now has to be in between 51 and 100. From this simple game we can adequately define information as that which reduces uncertainty.

Get access

Grade+20% off
$8 USD/m$10 USD/m
Billed $96 USD annually
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
40 Verified Answers

Related Documents

Biology 2581B Lecture Notes - Cytosine, Wild Type, Erwin Chargaff
scarletrabbit663
Biology 2581B Lecture Notes - Lecture 1: Guanine, Adenine, Cytosine
apricotsea-lion820
Biology 2581B Lecture 2: Lecture 2 – DNA.docx
cerisellama481

Related Questions

GENETICS QUESTION: For this experiment we focused on forward genetics: First, his- mutants were identified and then complementation analysis was used in order to determine which of the 7 his genes was mutated. ( The exact data can be seen at the bottom). In this experiment we screened for his- mutants that were generated via a random mutation process resulting from errors in DNA replication or induced by EMS aka Ethyl methanesulfonate. We compared a mutagenized culture that was treated with EMS and an un-mutagenized culture that was not treated with EMS. However, both cultures went trhough an enrichment process to increase the frequency of his- mutants.

Point mutations involve the substitution of one type of base pair for another. If the substitution replaces a purine with a purine (or a pyrimidine with a pyrimidine), it is called a transition. If the substitution replaces a purine with a pyrimidine (or a pyrimidine with a purine), it is called a transversion. Many mutagenic chemicals cause this type of mutation by alkylation of a nucleotide base. The chemically altered base can then be misread during replication or repair. For example the mutagen, ethyl methanesulfonate (EMS), adds an ethyl (-CH2-CH3) group to both guanine or thymine bases causing the modified guanine to pair with thymine and the modified thymine to pair with guanine.

1. Why might one gene be more susceptible to EMS mutagenesis than another gene?

2. Why might we have expected to find a higher frequency of polar mutants (his4A-B-C-) in the unmutagenized culture compared to the mutagenized culture?

Table 2. Experiment 1B class data summary (number of his-mutants by section)

Unmutagenized

Section

1

2

3

4ABC

4A

4B

4C

5

6

7

Total his- mutants

Tuesday

7

1

5

4

2

1

9

2

3

1

35

Wednesday

2

1

14

0

4

2

2

4

8

2

39

Total

9

2

19

4

6

3

11

6

11

3

74

% each mutant

12.1

2.7

25.7

5.4

8.1

4.1

14.9

8.1

14.9

4.1

~100%

Mutagenized

Section

1

2

3

4ABC

4A

4B

4C

5

6

7

Total his- mutants

Tuesday

2

5

4

2

6

5

5

0

4

1

34

Wednesday

9

1

15

0

9

1

2

2

9

4

52

Total

11

6

19

2

15

6

7

2

13

5

86

% each mutant

12.8

7.0

22.1

2.3

17.4

7.0

8.1

2.3

15.1

5.8

~100%

For years we have been able to analyze the fossils of extincthumans like those of Neanderthals. Their bones gave us clues to howthey looked, how they lived, and how they might be related to us.In the last few years, we have been able to get a better view intothese extinct species and their relationship to our species. Thanksto next generation genomic sequencing technologies, we have beenable to obtain genetic data on Neanderthals, Denisovans and othersthat have allowed us to make direct comparisons between our speciesand theirs. In addition, we have been able to observe the presenceof DNA from these extinct species within our own genomes which tellus that we had kids with these now extinct humans.

If we take Neanderthal DNA as an example, we find that we share99.5 percent of our DNA with them. Which means that we andNeanderthals were separated from each other for more than 500,000years. This also means that we were genetically similar enough tobe able to have offspring with each other. We see evidence of thisgenetic introgression between us and Neanderthals when we look atour DNA. Interestingly, if we obtain one thousand DNA samples fromGrossmont College students, we estimate that we would be able toextract at least 20% of the Neanderthal genome from them. Thismeans that 20% of our collective nucleotide sequences are sharedwith Neanderthals but this doesn’t mean that 20% of our genes areNeanderthal specific genes. Of the twenty-one thousand genes thatNeanderthals had, only a few dozen of their genes survive withinour genome. The same holds true for Denisovans. What is moreinteresting is that older versions of ourselves have a tendency tohave higher levels of DNA from extinct humans compared toourselves. In other words, over the course of thousands of years wehave been slowly weeding out the DNA of the other humans from ourgenome in favor of our genes.

One basic question that we might like to ask is why have we beenlosing Neanderthal and Denisovan genes over time? Also, why do westill have a significant amount of DNA segments that don’t seem tobe going anywhere? The answer to this has to do with selection.

The answer to the first part of the question has to do with thefact that we were not fully genetically compatible with the otherhuman species to begin with. Basically, to have had Neanderthal orDenisovan genes in the past meant that an individual's level offitness was reduced. To get rid of their genes in favor of our ownwould mean that the relative fitness of individuals would increaseovertime.

The answer to the second question has to do with the fact thatDNA segments from Neanderthals and Denisovans that did not code forproteins remained because there was little selective pressureagainst them. As far as genes are concerned, the only Neanderthaland Denisovan genes that remain only exist because they had aselective advantage or at least have been neutral to ourspecies.

DNA segments that that were neutral or that had a selectiveadvantage, whether genes or not, simply remained because they wereadvantages or there was little selective pressure against them.

Your task for this week is to look for information about oneNeanderthal or Denisovan gene that survive within our species andto tell me what function that gene has and how that gene mayincrease the fitness of modern human populations.

  • In your post, discussONE Neanderthal orDenisovan gene that is currently found within our species.
    • Tell us about the gene and its function. Explain what it doesfor us. (Make sure you give a citation.)
    • Explain in your own words any possible advantages that yourNeanderthal or Denisovan gene might have. Make sure you give ininformed opinion that is grounded in your research.

Introduction: A Chi-square test is used to compare observed data with expected data according to a hypothesis. For instance, if you were crossbreeding 2 heterozygous pea plants, you would expect to see a 3:1 phenotypic ratio in the offspring. In this case, if you were to breed 400 pea plants, you would expect to see 300 plants showing the dominant trait and 100 showing the recessive trait. But what happens if you observe only 260 plants with the dominant trait and 140 plants with the recessive trait? Does this mean something is wrong with Mendelian genetics or is this difference in expected results just due to chance (random sampling error)? These are the questions that can be answered using Chi-square statistics. The results of this statistical test is used to either reject or accept (fail to reject) the null hypothesis. The null hypothesis states there is no significant difference between the observed results and the expected results. This means that if the null hypothesis is accepted, the difference in observed and expected results was just a matter of chance and so the observed results basically "fit" with what was expected. Degrees of freedom (df) = number of independent outcomes (Y) being compared less 1 df = Y-1 At the 95% confidence interval we are 95% confident that there is a significant difference between the observed and expected results, therefore rejecting the null hypothesis. Probability Value - Is the decimal value determined from the X2 table and is the probability of accepting the null hypothesis. A 0.05 probability value equates to a 95% confidence interval.

The Chi-squared test formula is: Example: If we cross two pea plants that are heterozygous yellow pods, we would expect a 3:1 phenotypic ratio. So let's say we actually did the cross and got 280 plants with green pods and 120 plants with yellow pods. Question: Is this a 3:1 phenotypic ratio? This is the value of Chi-squared Test. We have a total of 400 plants and we expect a 300 green:100 yellow phenotypic ratio If the calculated Chi-squared value is less than the critical value listed in the Chi-squared table, then we accept the null hypothesis. This means that there is no significant difference between the observed and the expected values. Our degrees of freedom (df) = 2 outcomes - 1, or df = 1. Now we go the X2 table below and using the df = 1 and probability value of 0.05, our critical value is 3.84. Since our calculated X2 value is 5.33, and is larger than the critical value, we reject the null hypothesis and can say (at 95% confidence) that there is a significant difference between our observed and expected values.

The parent generation is yellowed podded and green podded pea plants. You cross a yellow podded pea plant with a green podded pea plant and you get 100% yellow podded plants in the F1 Generation (Phenotypic ratio 4 : 0, yellow to green). What will be the expected phenotypic ratio when you allow the F1 generation to reproduce?

Fill out the Punnett square.

If we actually did the cross and got 1150 yellow and 350 green. Would this be a consistent with what was expected?

Learning Outcomes Questions

1. Why would you run a Chi-squared test?

To determine if our data is consistent with expected results.

a To determine if our data is consistent with expected results.

b To determine if our data exactly matches the expected results.

c To determine the expected results.
d

To compare the phenotypic ratios to the genotypic ratios.

2. Determine the degrees of Freedom of the phenotypic ratio for this genetic cross.

a. 1

b. 2

c. 3

d. 4

e. 5

3. Using the data given, what is the result of your Chi-squared analysis? x2= ___.

a.

2.22
b

2.71
c

4.36
d

187.78
e

448.27

4. Using the results of your Chi-squared analysis, do we fail to reject or reject the null hypothesis?

a.

Fail to reject the null
b.

Reject the null
c.

It cannot be determined from the data given