welcome! week of february 16, 2009 lab topics –replica plating –meiosis problem solving...

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Welcome! Week of February 16, 2009 Lab Topics –Replica Plating –Meiosis Problem Solving –Meiosis –Binomial Probability –Chi-Square –Conditional Probability Reminders : •Save ALL graded docs & see p.11 •Exam THIS Thurs Due NEXT week: •Pre-lab quiz •Section assignment #2 •M&M Lab template •Problem Set

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Welcome!Week of February 16, 2009

•Lab Topics–Replica Plating–Meiosis

•Problem Solving–Meiosis–Binomial Probability–Chi-Square–Conditional Probability

Reminders:•Save ALL graded docs & see p.11•Exam THIS Thurs

Due NEXT week:•Pre-lab quiz

•Section assignment #2•M&M Lab template•Problem Set

Quiz & Revisit Last Week

• Quiz– 2:10-2:13

• Revisit – Assignment 1, #2c

GDP DisordersSign Up Sheets for:• Achondroplasia• Batten Disease• Polycystic Kidney Disease 1• Hemophilia A

• Check the 1B website for resources for your disorder• Use the OMIM # on the 1B website for your disease• Ernst Mayer Library (26 Oxford St—near MCZ)

– VERY eager to teach you how to use PubMed, OMIM, etc– 5 citation minimum. Must be peer-reviewed journal articles!

• See manual p.13-15, 20-22 for Pre-Draft 1 directions• Start EARLY!

Mutant Analysis, week 2

• What are we doing this week?

• Why are we doing this?

• Specifically, what is the experimental procedure for the lab?

• Hypothesize: What results do you expect the plates to look like next week?

Remember…• Keep hair/sleeves out of flame

• Sterile technique– Dishes closed, when not actively working– Bunsen burner—creates updraft – Use each toothpick ONE time– Do NOT flick lint off velvet

• After you clean up, work on, review meiosis problems (#1-7) in last week’s packet

Meiosis• For the top diagram:

– n=_____, 2n=_____

– What separates in anaphase I?

– What separates in anaphase II?

Mendel’s Laws & Meiosis• Segregation—2 alleles of a

gene segregate into different meiotic products

• Independent Assortment—unlinked genes are inherited (sorted) independently from each other

Crossing Over occurs in M1, forming tetrads

Mitosis vs MeiosisMitosis Meiosis

Results in 2 somatic (body) cells, each is 2n & identical to the parent cell

Produces 4 gametes--egg & sperm, each is 1n & different from each other and the 2n parent cell

Daughter cells for growth and repair Daughter cells for sexual reproduction

1 nuclear division: double genetic info of 2N cell, split the sisters

2 nuclear divisions: double genetic info of 2N cell, split the homologs, split the sisters

In metaphase chromosomes line up singlely

Homologs line up during M1 metaphase, creating a tetrad. Crossing over ensures genetic diversity

Lily Anther Lab--Meiosis

• Preparing lily anthers– Prepare several (3+) at a

time, to ensure you’ll find all stages of meiosis

• Divide & conquer– Left side focus on M1– Right side focus on M2

• Scopes– Do NOT use 100x--squish!– Once focused, do NOT use

coarse adjustment– Snap pictures of all stages– Upload AND name each

picture

Photos•Be selective

•Only upload one photo of each stage of M1 or M2

•Label photos•ie “JoeS1”, “JoeM2Meta”, etc

•Delete photos•From the desktop, iPhoto, and camera once uploaded

Discussion• Questions about the exam?

• Cheat Sheet Suggestions– Pattern of inheritance for diseases (PKU=

autosomal recessive)– M & M figures from lecture / textbook– Formulas– You do NOT need Chi Square table (p.141)

Elements of Probability

• Events: – A & A’ are complements– P{A} + P{A’} = 1

• 1- P{A’} = P{A}

• Compound Event– Union: P {A OR B}

• P{A+B} • Probability is bigger than the probability of just A or just B

– Intersection: P{A & B} • P{AB} • Probability is smaller; both A and B are less likely to occur

– P for snow = need both cold weather AND precipitation

• #8-10 in packet

Binomial Distribution• The binomial formula gives us a shorthand way to find the

probability of a group of mutually exclusive independent events (lots of coin tosses or lots of offspring).

– Where A&B are mutually exclusive events (ex boys vs girls)

– p = probability of A occurring in any one trial

– 1-p = probability of event B occurring in any one trial

– n = total # trials (eg, # offspring examined)

– r = # of “event A”s we are interested in

– n-r = # of “event B”s we are interested in

• Key phrase= “exactly n of x and m of y”• #11-14 in packet

Chi-Square Tests (#18-24)

• Goodness of fit test– Determines how well observed results “fit”

or agree with expected results• Obs: 651 A_ : 207 aa vs Exp : 3 A_:1 aa

• Homogeneity (Association) test– Determines how homogeneous (similar)

two observed (experimental) results are• Observe: 651 A_ : 207 aa and 787 A_ : 277 aa

Using Goodness of Fit(p.139 in text)

• State the null hyp (no difference between O and E)

• Use rules of probability to predict the types and proportions of progeny expected if hypothesis is true (Aa x Aa)

• Convert proportions and % to numbers

• Use the formula

Interpreting the x2 Value• Closer the x2 value is to 0 = closer O matches E• Calculate df: # classes data -1• Use graph (p.141) to determine P value

– Assuming the null hyp is true, probability that a worse or equally bad fit (as large/larger X2 value) would be obtained by chance

– If p>0.05, results are non-significant (small difference btn E & O), and we fail to reject the null hyp

– If p<0.05, results are significant (big diff btn E & O) & we reject the null hyp

Test of Homogeneity• State null hyp: both sets of data come from the same distribution

• Solve--I highly recommend the “plug and chug” method

• df always 1; comparing 2 groups of data (df = 2-1)

• Interpret p value– If p>0.05, results are non-significant (small difference btn E &

O), and we fail to reject the null hyp– If p<0.05, results are significant (big diff btn E & O) & we reject

the null hyp

Conditional Probability,Cookie Style

• Event A =• Event B =

• There are 2 bowls• Bowl #1

– 10 choc chip cookies– 30 plain cookies

• Bowl #2– 20 choc chip cookies– 20 plain cookies

• You pick a bowl at random & then a cookie at random. The cookie is plain.

• How probable is it that you picked from bowl #1, given it is plain?

Conditional Probability

What we’ve been calculating:

• We already know how to predict genetic ratios about future generations, given present info.– Aa x Aa =

• Conditional probability allows us:

• To infer genetic information about the past, given present info.

Conditional Probability,Cookie Style

• Event A = You pick bowl #1• Event B = You picked a plain

cookie.

• There are 2 bowls• Bowl #1

– 10 choc chip cookies– 30 plain cookies

• Bowl #2– 20 choc chip cookies– 20 plain cookies

• You pick a bowl at random & then a cookie at random. The cookie is plain.

• How probable is it that you picked it out of bowl #1, given it is plain?

Conditional Probability,Cookie Style

• Event A = You pick bowl #1• Event B = You picked a plain

cookie.• To compute P(A|B), we first

need to know:– P(A), the probability that you

picked bowl #1 (regardless of any other information)

– P(B), the probability of getting a plain cookie (regardless of any other information)

– P(B|A), the probability of getting a plain cookie given you picked bowl #1.

• There are 2 bowls• Bowl #1

– 10 choc chip cookies– 30 plain cookies

• Bowl #2– 20 choc chip cookies– 20 plain cookies

• You pick a bowl at random & then a cookie at random. The cookie is plain.

• How probably is it that you picked it out of bowl #1, given it is plain?

Conditional Probability,Cookie Style

• Event A = You pick bowl #1• Event B = You picked a plain

cookie.• To compute P(A|B), we first need

to know:– P(A) = 1/2 (only 2 bowls)– P(B) = 80 cookies, 50 plain =.625– P(B|A) = 40 cookies in bowl #1 and

30 of them are plain = 30/40 = .75

• There are 2 bowls• Bowl #1

– 10 choc chip cookies– 30 plain cookies

• Bowl #2– 20 choc chip cookies– 20 plain cookies

• You pick a bowl at random & then a cookie at random. The cookie is plain.

• How probably is it that you picked it out of bowl #1, given it is plain?

Conditional Probability,Cookie Style

• Event A = You pick bowl #1• Event B = You picked a plain

cookie.• To compute P(A|B), we first need to

know:– P(A) = .5 (only 2 bowls)– P(B) = 80 cookies, 50 plain = .625– P(B|A) = 40 cookies in bowl #1 and

30 of them are plain = 30/40 = 0.75• Plug & Chug!

• There are 2 bowls• Bowl #1

– 10 choc chip cookies– 30 plain cookies

• Bowl #2– 20 choc chip cookies– 20 plain cookies

• You pick a bowl at random & then a cookie at random. The cookie is plain.

• How probably is it that you picked it out of bowl #1, given it is plain?

Conditional Probability,Cookie Style

• Event A = You pick bowl #1• Event B = You picked a plain

cookie.• To compute P(A|B), we first need to

know:– P(A) = ½ = .5 (only 2 bowls)– P(B) = 80 cookies, 50 plain = .625– P(B|A) = 40 cookies in bowl #1 and

30 of them are plain = 30/40 = 0.75• Plug & Chug!

• There are 2 bowls• Bowl #1

– 10 choc chip cookies– 30 plain cookies

• Bowl #2– 20 choc chip cookies– 20 plain cookies

• You pick a bowl at random & then a cookie at random. The cookie is plain.

• How probable is it that you picked it out of bowl #1, given it is plain?

Problems to Try

• #10, 11 in packet