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Molecular Biology and Genetics > Lecture 11 > Flashcards

Lecture 11 Flashcards

Mendelian Genetics

1
Q

RNA- Induced Transcriptional Silencing (RITS) in Yeast

A
  • 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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2
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RNAi as a Research Tool

A
  • 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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3
Q

Mendel’s model organism: pea plant

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

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4
Q

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

A
  • 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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5
Q

Artificial cross-fertilization of pea plants

A
  • 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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6
Q

Life Cycle of The Pea Plant

A
  • 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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7
Q

Mendel’s Hypotheses

A
  • 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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8
Q

A note about Nomenclature…

A

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

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9
Q

Mendel’s Approach Followed the Modern Scientific Method

A
  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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10
Q

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

A
  • 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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11
Q

Test Mendel’s Hypotheses

A
  • 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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12
Q

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

A

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

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13
Q

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

A

It will not produce any white flowers. (ww)

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14
Q

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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15
Q

Replicate-, Reciprocal-, and Test Cross Analysis

A
  • 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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16
Q

2.3 Dihybrid and Trihybird Crosses Reveal the Independent Assortment of Alleles

A
  • 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
Q

Dihybrid-Cross Analysis of Two Genes

A
  • 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
Q

An Aid to Prediction of Gamete Frequency

A
  • 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
Q

Mendel’s Second Law

A
  • 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
Q

Independent Assortment of Alleles from the RrGg x RrGg Cross

A
  • 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
Q

Trihybrid Crosses (and beyond)

A
  • 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
Q

Trihybrid Crosses : a sample prediction

A
  • 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
Q

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

A
  • 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
Q

The Probability that particular outcomes are consistent with the expected.

A
  • 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
Q

The probability that particular outcomes are consistent with the expected

A
  • 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
Q

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

A
  • 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
Q

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

A
  • 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
Q

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

A
  • 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
Q

Binomial Probability

A
  • 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
Q

Construction of a Binomial Expansion Formula

A
  • 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
Q

Binomial Expansion Formula – Example

A
  • 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
Q

Using Pascal’s triangle of binomial coefficients to solve: 

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

A

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.

Molecular Biology and Genetics (5 decks)

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