Log in
HomeClass Notes1,200,000CA670,000

BI110 Lecture Notes - Lecture 3: E-Book, Pigment

11 views8 pages
rubymoose317
11 Nov 2015
School
WLU
Department
Biology
Course
BI110
Professor
Matthew Smith

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

Lecture 3 lec 3 page 1 lec 3 page 2. The liken that helped light colored moths thrive was killed. This changed the relative population dynamics the dark color moths become more prevalent. Both require genetic diversity macroscale often results in two species due to factors like geographic isolation. This happens over much longer spans of time. lec 3 page 3 lec 3 page 4. Chemical energy and matter are converted to do work. Defined by the wave length = the difference between two peeks or vally"s. Amount of energy is inversely proportional to wave length. When photon encounters matter one of three things happen: Absorption of light is most important for forms of life. This means at the atomic level is the electrons are delocalized. They are more avalible to absorb photons of light lec 3 page 5 lec 3 page 6. Red promote from ground state to energy level one.

Get access

Grade+
$40 USD/m
Billed monthly
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
10 Verified Answers
Class+
$30 USD/m
Billed monthly
Class+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
7 Verified Answers

Related Documents

BI110 Lecture Notes - Lecture 6: Covalent Bond, Base Pair, Uracil
greenmammoth432
BI110 Lecture Notes - Lecture 1: Industrial Revolution, Soot, Genetic Diversity
whitefish372
BI110
Final Exam
Study Guide
BI110- Final Exam Guide - Comprehensive Notes for the exam ( 99 pages long!)
ceriserhinoceros44

Related Questions

A geneticist studying agriculturally important traits inanimals, designs an experiment in chickens to determine the numberof genes involved in determining the number of eggs layed permonth. A variety of chicken that lays 120 eggs per month is crossedto a variety of chicken that lays 200 eggs per month. Theiroffspring (F1 generation) lay 160 eggs per month. TheF2 egglaying phenotypes vary greatly over a range withthe highest and lowest equal to the parental extremes. A total of13,000 F2 offspring are produced with 12 laying 120 eggsper month and 13 laying 200 eggs per month.

  1. How many gene pairs are involved in determining the number ofeggs a chicken lays? See the last page for a tutorial oncalculating number of polygenes.
  2. What is the average contribution, in eggs, of each activeallele? _______________

Additional information for the polygene question.

To calculate the number of polygenes: 1/4n = ratio ofF2 individuals expressing either extreme phenotype. In the examplegiven in your textbook and mini-lecture of wheat kernel color, 1/16of the progeny were either dark red or white like the P generation.You need to solve the equation 1/4n = 1/16. n = thenumber of genes. In this example n=2, there are two genes involvedin wheat kernel color.

Determination of the Number of Polygenes (n) involved in aQuantitative Trait

n

Individuals expressing either extremephenotype

Distinct phenotypic classes

1

1/4

3

2

1/16

5

3

1/64

7

4

1/256

9

5

1/1024

11

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%

1. Codominance results in:

a) joint expression of two alleles
b) intermediate expression of two alleles
c) sometimes dominance , sometimes recessiveness
d) none of the above

(I am stuck between A or B. Why is A or B the answer? TYSM!)

CAN you please explain how to get these two following questions? I am kind of confused. TYSM!-->

2) In drosophila, fused veins is inherited as an X-linked recessive. If a fused vein male is crossed with a wild type female, the phenotypes of the F1 offspring will be:

A) All fused vein
B) All wild type
c) 1/2 wild type females, 1/2 fused vein males
D) 1/2 fused vein females, and 1/2 wild type males
E) 1/2 wild type famels: 1/4 wild type males: 1/4 fused vein males

3) The recessive muations curved wing and sepai eye are located on different autosomal chromosomes in drosophila. A curved wing female is crossed to a sepia eye male to produce the F1. If an F1 female is then crossed to a curved wing, sepia male, what phenotypic ratio will be expected in their offspring?

A) 1/4 WT ; 1/4 sepia; 1/4 curved; 1/4 sepia, curved
B) 2/4 WT; 1/4 Sepia; 1/4 curved
C) 9/16 WT; 3/16 sepia; 3/16 curved; 1/16 sepia, curved
D) 1/2 sepia; 1/2 curved