Flashcards in Lecture 11 Deck (32)

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1

## RNA- Induced Transcriptional Silencing (RITS) in Yeast

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- Is responsible for heterochromatin formation at centromere.

- RNA Pol II transcription at the centromere occurs during S-Phase of cell cycle.

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## RNAi as a Research Tool

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- RNAi can be used in a multitude of ways in research

- RNAi can be used to selectively knock down expression of selected genes to determine the effect on the phenotype.

- It may also be effective in medicine, where it might be used to control expression of genes that produce too much transcript, such as in cancer, or produce abnormal transcripts.

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## Mendel's model organism: pea plant

### Fertilization -> Seed development -> Mature seeds/ Germination -> Plant Growth -> Plant Maturation, flower development.

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## Gregor Mendel brought his experience in other scientific fields to questions he asked about heredity.

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- Physics: Rock is not a big solid homogeneous object- is actually made of tiny (invisible) particles of matter...

- Math: the frequency of chance events (like flipping of a coin) can be predicted, using mathematical equations. Probability theory.

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## Artificial cross-fertilization of pea plants

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- Emasculate purple flowers by removing anthers (male)

- Transfer pollen from white flower anthers (male) to purple flower ovule (female)

- Fertilization occurs

- Seeds develop

- Seeds planted, plants grow, and traits recorded.

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## Life Cycle of The Pea Plant

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-Fertilization, Seed Development, Germination, Plant Growth, Plant Maturation, flower development.

- Generate "true-breeding" plants for dichotomous traits.

- Dichotomous traits = 2 alternatives ONLY.

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## Mendel's Hypotheses

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-Hypothesis: Dichotomous trait (2 alternatives) like purple/white color of flowers is the result of two alternative forms (alleles) of a "particle of heredity" (gene). In a pure breeding purple plant, both alleles are WW; in apure breeding white plant, both alleles are ww.

- Observation: F1 plants are all purple.

-Hypothesis: one allele is DOMINANT over the other

- Genotype of F1 =W(purple) w( white)

-Ww is purple. Implication: purple (W) is dominant to white (w).

-Observation: F2 plants are 3 purple to 1 white

- Hypothesis: independent segregation of alleles in germ cells and "chance" combinations of alleles in progeny.

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## A note about Nomenclature...

### Ww x Ww is known as "monohybrid" cross: both parents are heterozygous (hybrid) for the same, one (mono) trait.

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## Mendel's Approach Followed the Modern Scientific Method

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1. Make initial observations about a phenomenon or process.

2. Formulate a testable hypothesis.

3. Design a controlled experiment to test the hypothesis.

4. Collect data from the experiment.

5. Interpret the experimental results, comparing them to those expected under the hypothesis.

6. Draw a conclusion and reformulate the hypothesis if necessary.

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## Test Mendel's Hypotheses: In this punnett square, the mendelian hypotheses are presented in table form.

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- Hypothesis: pollen grains each have only one allele (segregating) and the likelihood of having the W allele is equal to the likelihood of having the w allele (same for the egg)

- Hypothesis: an allele from pollen joins with an allele from egg "randomly", the frequency of any particular combination is dictated by the frequency of each allele = PRODUCT of the two frequencies.

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## Test Mendel's Hypotheses

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-Phenotypes: 3 purple plants to every one white plant = observations consistent with hypothesis!

- Genotypes: 1WW : 2Ww : 1ww. How to confirm this?

- But all that Mendel can see is the color of the flowers (phenotype). So...

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## How to prove that some of the purple plants are WW (homozygous) and some are Ww (heterozygous)?

### -TEST CROSS: Cross each purple (W_) F2 plant to a true breeding white plant (ww).

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## Think about it: Why is a cross to the homozygous dominant plant (WW) NOT a "test cross"?

### It will not produce any white flowers. (ww)

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## Mendel's "law of segregation"

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In SUMMARY:

- The two alleles for each trait will separate (segregate) from one another during gamete formation, and each allele will have an equal probability (1/2) of inclusion in a gamete.

- Random union of gametes at fertilization will unite one gamete from each parent to produce progeny in ratios that are determined by chance = probability theory.

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## Replicate-, Reciprocal-, and Test Cross Analysis

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- Mendel made many replicate crosses, producing hundreds or thousands of progeny, repeating each cross multiple times.

- He performed reciprocal crosses, in which the same genotypes are crossed, but the sexes of the parents are reversed.

- He also performed text crosses (as just described)

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## 2.3 Dihybrid and Trihybird Crosses Reveal the Independent Assortment of Alleles

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- Each of the seven traits Mendel studied showed the same pattern of heredity, explained by the Law of Segregation.

- Mendel also studied the inheritance of two or more traits simultaneously

- This work lead to Mendel's Law of Independent Assortment (Mendel's Second Law).

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## Dihybrid-Cross Analysis of Two Genes

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- To study the simultaneous transmission of two traits, Mendel made dihybrid (vs monohybrid) crosses = crosses between organisms that differed for two traits.

- He began each cross with pure breeding lines (RRGG and rrgg) and produced F1 that were heterozygous for both traits (RrGg)

- See (pea) phenotype

a. Rr (named for recessive allele =rugous =wrinkled). Dichotomous : round or wrinkled.

b. Gg (named for recessive allele = green.) Dichotomous: yellow or green.

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## An Aid to Prediction of Gamete Frequency

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- The forked-line diagram is used to determine theoretical gamete genotypes and frequencies.

- In a hybrid individual, if alleles for two different genes segregate independently of one another = independent assortment, then predict 4 different genotypes in the GAMETES produced.

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## Mendel's Second Law

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- Mendel's law of independent assortment

- During gamete formation, the segregation of alleles at one gene is independent of the segregation of alleles at another gene.

- In an RrGg individual, for example, expect RG, rG, Rg, and rg gametes to form in equal numbers.

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## Independent Assortment of Alleles from the RrGg x RrGg Cross

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- Instead of using Punnett Square, you can use monohybrid 3:1 dominant:recessive phenotype ratio expected in F2 to PREDICT phenotype ratios in progeny from a dihybrid (RrGg x RrGg) cross.... 3/4 dominant phenotype; 1/4 recessive phenotype.

- round, yellow (R_G_) (3/4)(3/4) = 9/16

- round, green (R_gg) (3/4)(1/4) = 3/16

- wrinkled, yellow (rrG_) (1/4)(3/4) = 3/16

- wrinkled, green (rrgg) (1/4)(1/4) =1/16

- Probability rule: the probability of two independent events taking place simultaneously is the PRODUCT of each probability.

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## Trihybrid Crosses (and beyond)

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- The number of gamete genotypes can be expressed as 2n where n = number of genes and 2, because each trait is dichotomous. (ex. 3 traits= 2 x 2 x 2 = 8 gamete genotypes from a trihybrid)

- Possible phenotypes in a trihybrid cross?

- 8 male gamete genotypes, randomly fused to 8 female gamete genotypes = 64 different progeny.

-But, some genotypes are "repeats" for phenotype -look at Punnett square)

- Using the 3/4, 1/4 expected frequencies for each individual trait, a phenotypic ratio expected for F2 progeny can be generated.

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## Trihybrid Crosses : a sample prediction

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- Example: Crossing RrGgWw x RrGgWw.

- How many progeny will have the phenotype: wrinkled and Yellow seeds, and white flowers?

- First consider, what genotypes will give rise to this phenotype?

- rrG_ww

-What is the predicted frequency of this combination of phenotypes? Remember: 1/4 RR, 1/2 Rr, 1/4 rr.

- rr = 1/4; G_ = Gg + GG =3/4; ww = 1/4.

So... 1/4 x 3/4 x 1/4 = 3/64.

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## 2.5 Chi-Square Analysis Tests the Fit between Observed and Expected Outcomes

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- Scientists must be able to make comparisons and expected results to objectively determine whether results are consistent with expectations.

- The chi-square test was developed to allow for these objective comparisons.

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## The Probability that particular outcomes are consistent with the expected.

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- The chi-square test is commonly used for quantifying how closely an experimental observation matches the expected outcome. The null hypothesis is that they MATCH.

- More formally, the null hypothesis is that any difference you see between the observed vs expected is small and due to sampling chance -- not due to real difference.

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## The probability that particular outcomes are consistent with the expected

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- First, calculate the amount of deviation from the expected.

- χ2 = ∑(O − E)^2/E

- The bigger the differences, the bigger the X^2, the less likely that the observed is close to the expected and is just differing from it "by chance".

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## The probability that particular outcomes are consistent with the expected (part 2)

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- Next, interpret the observed differences

- Null Hypothesis: observed ~ expected (slight differences in value due to chance).

- Probability (P) value. How probable is it that this null hypothesis is correct?

- If p > 0.05, accept the null hypothesis

- If p< 0.05, reject the null hypothesis.

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## X^2 test: Sometimes we are hoping p > 0.05 (want observed to match predicted)

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- In the case of Mendel testing an hypothesis:

-Observed values equal the expected values = null hypothesis

- Hypothesis Mendel used to derive "expected" is SUPPORTED if the observed and expected are the same.

- If obtain a very small p-value (less than 0.05), then "observed" are NOT as "expected" (ARE NOT the same). Mendel goes back to the drawing board to come up with another method of inheritance.

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## Sometimes we are hoping p < 0.05 ( want to prove that an intervention has had a real effect)

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-In the case of developing a therapy that will kill cancer cells, for example:

- Untreated cancer cells vs treated cancer cells, asking if the two behave the same way - null hypothesis is that they do.

- If obtain a very small p-value (less than 0.05), then null hypothesis is REJECTED. Treated and untreated ARE NOT behaving the same way. Researcher is encouraged- implies that the treatment is having an effect.

- If obtain a p-value > 0.05, then the small difference you are seeing before and after treatment is not significant - back to the drawing board...

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## Binomial Probability

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- Some genetic questions involve predicting the likelihood of a series of events (for which there are two or more possible outcomes each time)

- We use binomial probability calculations to answer this type of question.

- For families with three children, predict the proportions for each possible combination of boys and girls.

GGB

GBB

BBB

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## Construction of a Binomial Expansion Formula

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- A binomial expansion contains two variables; p, the frequency of one outcome, and q, the frequency of the alternative outcome (p and q may or may not be equal, depending on the type of outcome.)

- (p+q) = 1, since these are the only two outcomes.

- We expand the equation by the power of n, where n= the number of successive events: (p+q)^n.

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## Binomial Expansion Formula -- Example

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- For families with three children, predict the proportions for each possible combination of boys and girls.

- p = probability of a boy =1/2.

- q = probability of a girl = 1/2

-(note: law of segregation of XY in gametes of father)

- Binomial Expansion: (p + q)^3 = p^3 + 3p^2q + 3pq^2 + q^3 = 1

- p^3 = ½ x ½ x ½ = 1/8 (3 boys); 3p^2q = 3 (½ x ½ x ½) = 3/8 (2 boys, 1 girl); 3pq^2 = 3/8 (1 boy, 2 girls); q^3 = 1/8 (3 girls) = 1

- The math makes sense, but what is the conceptual basis for the math?

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