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1

RNA- Induced Transcriptional Silencing (RITS) in Yeast

- Is responsible for heterochromatin formation at centromere.
- RNA Pol II transcription at the centromere occurs during S-Phase of cell cycle.

2

RNAi as a Research Tool

- 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.

3

Mendel's model organism: pea plant

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

4

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

- 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.

5

Artificial cross-fertilization of pea plants

- 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.

6

Life Cycle of The Pea Plant

-Fertilization, Seed Development, Germination, Plant Growth, Plant Maturation, flower development.
- Generate "true-breeding" plants for dichotomous traits.
- Dichotomous traits = 2 alternatives ONLY.

7

Mendel's Hypotheses

-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.

8

A note about Nomenclature...

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

9

Mendel's Approach Followed the Modern Scientific Method

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.

10

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

- 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.

11

Test Mendel's Hypotheses

-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...

12

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).

13

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)

14

Mendel's "law of segregation"

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.

15

Replicate-, Reciprocal-, and Test Cross Analysis

- 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)

16

2.3 Dihybrid and Trihybird Crosses Reveal the Independent Assortment of Alleles

- 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).

17

Dihybrid-Cross Analysis of Two Genes

- 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.

18

An Aid to Prediction of Gamete Frequency

- 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.

19

Mendel's Second Law

- 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.

20

Independent Assortment of Alleles from the RrGg x RrGg Cross

- 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.

21

Trihybrid Crosses (and beyond)

- 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.

22

Trihybrid Crosses : a sample prediction

- 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.

23

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

- 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.

24

The Probability that particular outcomes are consistent with the expected.

- 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.

25

The probability that particular outcomes are consistent with the expected

- 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".

26

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

- 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.

27

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

- 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.

28

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

-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...

29

Binomial Probability

- 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

30

Construction of a Binomial Expansion Formula

- 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.

31

Binomial Expansion Formula -- Example

- 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?

32

Using Pascal’s triangle of binomial coefficients to solve: 

what % pods will have 3 yellow and 3 green seeds?

Conclusion: In a monohybrid cross for yellow/green seed color, you can expect 13% of the pods you recover to have 3 yellow and 3 green seeds.