Lecture 11

Question 1

RNA- Induced Transcriptional Silencing (RITS) in Yeast

Answer 1

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

Question 2

RNAi as a Research Tool

Answer 2

• 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.

Question 3

Mendel’s model organism: pea plant

Answer 3

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

Question 4

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

Answer 4

• 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.

Question 5

Artificial cross-fertilization of pea plants

Answer 5

• 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.

Question 6

Life Cycle of The Pea Plant

Answer 6

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

Question 7

Mendel’s Hypotheses

Answer 7

• 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.

Question 8

A note about Nomenclature…

Answer 8

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

Question 9

Mendel’s Approach Followed the Modern Scientific Method

Answer 9

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.

Question 10

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

Answer 10

• 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.

Question 11

Test Mendel’s Hypotheses

Answer 11

• 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…

Question 12

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

Answer 12

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

Question 13

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

Answer 13

It will not produce any white flowers. (ww)

Question 14

Mendel’s “law of segregation”

Answer 14

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.

Question 15

Replicate-, Reciprocal-, and Test Cross Analysis

Answer 15

• 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)

Question 16

2.3 Dihybrid and Trihybird Crosses Reveal the Independent Assortment of Alleles

Answer 16

• 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).

Question 17

Dihybrid-Cross Analysis of Two Genes

Answer 17

• 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.

Question 18

An Aid to Prediction of Gamete Frequency

Answer 18

• 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.

Question 19

Mendel’s Second Law

Answer 19

• 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.

Question 20

Independent Assortment of Alleles from the RrGg x RrGg Cross

Answer 20

• 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.

Question 21

Trihybrid Crosses (and beyond)

Answer 21

• 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.

Question 22

Trihybrid Crosses : a sample prediction

Answer 22

• 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.

Question 23

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

Answer 23

• 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.

Question 24

The Probability that particular outcomes are consistent with the expected.

Answer 24

• 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.

Question 25

The probability that particular outcomes are consistent with the expected

Answer 25

• 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”.

Question 26

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

Answer 26

• 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.

Question 27

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

Answer 27

• 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.

Question 28

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

Answer 28

• 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…

Question 29

Binomial Probability

Answer 29

• 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

Question 30

Construction of a Binomial Expansion Formula

Answer 30

• 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.

Question 31

Binomial Expansion Formula – Example

Answer 31

• 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?

Question 32

Using Pascal’s triangle of binomial coefficients to solve: 

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

Answer 32

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.