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Announcements. Please don’t interrupt other classes (including other Genetics labs) to check flies in Brooks 204 (see schedule on the door). Bring calculators for next week’s lab. Homework: practice on problems 6,7,8,12 - do not turn in.

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slide1
Announcements
  • Please don’t interrupt other classes (including other Genetics labs) to check flies in Brooks 204 (see schedule on the door).
  • Bring calculators for next week’s lab.
  • Homework: practice on problems 6,7,8,12 - do not turn in.
          • Turn in answers to problem set 1, next page, in lab next week.
slide2
Problem Set 1: due 9/10, 9/11 in lab

1. In guinea pigs, rough coat (R) is dominant over smooth coat (r). A rough coated guinea pig is bred to a smooth one, giving eight rough and seven smooth progeny in the F1.

What are the genotypes of the parents and their offspring?

If one of the rough F1 animals is mated to its rough parent, what progeny would you expect?

In summer squash, white fruit (W) is dominant over yellow (w), and disk-shaped fruit (D) is dominant over sphere-shaped fruit (d). The following problems give the phenotype of the parents and their offspring. Determine the genotypes of the parents in each case:

White, disk x yellow, sphere gives 1/2 white, disk and 1/2 white, sphere.

White, sphere x white, sphere gives 3/4 white, sphere and 1/4 yellow, sphere.

Yellow, disk x white, sphere gives all white, disk progeny.

2 pts each answer for total of 10 pts

review of last lecture
Review of last lecture

Regulation of the cell cycle - 3 main checkpoints

Meiosis: 2 sequential divisions: one reductional,

one equational. Sources of genetic variation: recombination via crossing over in pachytene of prophase I and independent assortment, metaphase I.

Mendel and his peas - brief intro

Monohybrid cross - Punnett square method

slide4
Monohybrid Cross:

Punnett Square Method

  • (1) Define symbols:
    • D = tall allele
    • d = dwarf allele
  • (2) State the cross
  • (3) Diagram the gametes
  • (4) Complete the squares
  • (5) Summarize the results:
    • Genotype
    • Phenotype
slide5
Reciprocal crosses
  • Results were the same regardless of which parent was used, e.g.
    • tall pollen pollinating dwarf eggs
    • dwarf pollen pollinating tall eggs
  • Therefore the results were not sex-dependent
  • Mendel proposed “unit factors” to explain his results
outline of lecture 4
Outline of Lecture 4
  • Mendel’s first three postulates
  • Monohybrid Testcross
  • Dihybrid Cross

IV. Independent Assortment

  • Trihybrid Cross
  • Molecular basis of Mendel’s postulates
  • VII. Probability

Basic Mendelian genetics

slide7
I. Mendel’s postulates

Postulate 1. Unit factors in pairs

  • Genetic characters are controlled by unit factors in pairs.
  • In other words, genes are present in two associated copies in diploid organisms.
  • For example, DD plants have two alleles for tallness, dd plants have two alleles for dwarfism.
slide8
Postulate 2. Dominance/recessiveness
  • In the case of unlike unit factors, one can bedominant and the other can berecessive.
  • In other words, when two different alleles of a gene are present, one may show its effect while the other may be masked.
  • For example, Dd plants have a tall allele D and a dwarf allele d, but are phenotypically tall.
slide9
Postulate 3. Segregation
  • During the formation of gametes, unit factors segregate randomly.
  • In other words, when sperm and eggs are formed, one of each allelic pair is randomly distributed to to each gamete.
  • For example, a Dd plant makes pollen or eggs, each randomly receives either the D allele or the d allele.
slide10
Practice: Axial/Terminal Pods
  • In garden peas, an allele T for axial flowers is dominant to an allele t for terminal flowers.
    • In the F2 generation of a monohybrid cross, what is the expected ratio of axial : terminal?
    • Among the F2 progeny, what proportion are heterozygous?
    • Among the F2 progeny with axial flowers, what proportion are heterozygous?
slide11
II. Monohybrid Test Cross (Backcross)
  • How can you determine genotype from individual expressing dominant phenotype? - DD or Dd?
  • Cross individual with dominant phenotype to a homozygous recessive individual.

?

?

slide12
III. Dihybrid cross - phenotypes

1.Which traits are dominant?

2. Did the phenotypes of the P1 generation affect the F1 or F2 generations?

slide13
Analysis of dihybrid cross phenotypes

(forked-line/probability method)

Trait 1

Trait 2

Combined traits

since yellow and round are dominant let g yellow g green w round w wrinkled
Since yellow and round are dominant,Let G = yellow, g = green, W = round, w = wrinkled.

Dihybrid Cross: P1 cross

Confirm on your own using a Punnett square!

slide17
Dihybrid Testcross:How to determine the genotype of an individual with 2 traits of dominant phenotype

All yellow

All Round

Mixed

slide18
IV. Independent assortment

Mendel’s Fourth Postulate: An Interpretation from the Dihybrid cross

  • During gamete formation, segregating pairs of unit factors assort independently.
  • In other words, segregation of 2 alleles at one genetic locus has no effect on the segregation of 2 alleles at another locus (unless linked).
  • For example, the assortment of yellow and green alleles has no effect on the assortment of round and wrinkled alleles, and vice versa.
slide19
Independent Assortment
  • Results in extensive genetic variation
  • Number of possible gametes = 2n where n is the haploid number
  • For humans, 223 = 8 million
  • Each individual represents one of (8 X 106)2 = 64 X 1012 possible genetic combinations from her parents
slide20
V. Gamete Formation: Trihybrid cross

What size of Punnett square needed for analysis?

slide22
VI. Molecular Basis of Mendel’s Postulates: Chromosome behavior
  • 1879: Walter Flemming discovers chromosomes in living cells.
  • 1900: DeVries, Correns, and Tschermak repeat, rediscover Mendel.
  • 1902: Sutton and Boveri and others link behavior of chromosomes to Mendelian segregation and independent assortment; propose chromosomal theory of heredity.
slide23
Theodor Boveri
  • Hypothesis: Removal of chromosomes should result in some change to organism.
  • Did this in sea urchin embryos, saw abnormal embryos.

1862-1915

slide24
Walter Sutton (1877-1916)
  • Studied grasshopper testis and ovary cells, observed that haploid numbers of chromosomes mixed to form diploid set.
  • Noted that this chromosome behavior correlated with Mendel’s genes, and proposed that genes are carried on the chromosomes.
slide25
Correlation Between Unit Factors and Genes on Chromosomes
  • Unit factors in pairs ~ genes on homologous chromosomes in pairs
  • Segregation of unit factors during gamete formation ~ genes on homologs segregate during meiosis
  • Independent assortment of segregating unit factors ~ genes on nonhomologous chromosomes assort independently
  • Stronger evidence for the chromosomal theory of heredity came from experiments of T.H. Morgan and others with fruit flies from 1909 onwards.
slide26
VII. Laws of probability help explain

genetic events

Genetic ratios are most properly expressed as probabilities:

ex. 3/4 tall: 1/4 dwarf

The probability of each zygote having the genetic potential for becoming tall is 3/4, etc..

Probabilities range from 0 (an event is certain NOT to happen), to 1.0 (an event is certain to happen).

How do we calculate the probability of 2 or more events happening at the same time?

slide27
Product law
  • For simultaneous outcomes (AND)
  • What is the chance that you will roll snake eyes with two dice? (1 and 1)
    • Chance of rolling 1 with first die = 1/6
    • Chance of rolling 1 with second die = 1/6
    • Chance of rolling two 1’s = 1/6 X 1/6 = 1/36
  • We used product law when calculating probabilities by the forked-line method.
slide28
Sum law
  • For outcomes that can occur more than one way (OR)
  • What is the chance that you will roll either a 1 or a 6 with one die?
    • Chance of rolling 1 = 1/6
    • Chance of rolling 6 = 1/6
    • Chance of rolling 1 or 6 = 1/6 + 1/6 = 2/6 = 1/3
slide29
Genetic Example

What is the probability (p) of green, wrinkled seeds (ggww) from a dihybrid cross?

Parents: GgWw X GgWw

Possible gametes: GW, Gw, gW, gw so p(gw)= 1/4

One gw must come from each parent (gw and gw), so p(ggww) = p(gw and gw) = 1/4 X 1/4 = 1/16

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