Announcements Please dont interrupt other classes (including other

Announcements Please dont interrupt other classes (including other

Announcements Please dont interrupt other classes (including other Genetics labs) to check flies in Brooks 204 (see schedule on the door). Bring calculators for next weeks 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. 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. a) What are the genotypes of the parents and their offspring? b) If one of the rough F1 animals is mated to its rough parent, what progeny would you expect? 2. 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: a) White, disk x yellow, sphere gives 1/2 white, disk and 1/2 white, sphere.

b) White, sphere x white, sphere gives 3/4 white, sphere and 1/4 yellow, sphere. c) Yellow, disk x white, sphere gives all white, disk progeny. 2 pts each answer for total of 10 pts Review of last lecture 1. Regulation of the cell cycle - 3 main checkpoints 2. 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. 3. Mendel and his peas - brief intro 4. Monohybrid cross - Punnett square method 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 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 I. Mendels first three postulates

II. Monohybrid Testcross III. Dihybrid Cross Basic Mendelian genetics IV. Independent Assortment

V. Trihybrid Cross VI. Molecular basis of Mendels postulates VII. Probability I. Mendels 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. Postulate 2. Dominance/recessiveness In the case of unlike unit factors, one can be dominant and the other can be recessive. 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. 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. 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? 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. ?

? III. Dihybrid cross - phenotypes 1.Which traits are dominant? 2. Did the phenotypes of the P1 generation affect the F1 or F2 generations? Analysis of dihybrid cross phenotypes

(forked-line/probability method) Trait 1 Trait 2 Combined traits Dihybrid Cross: P1 cross Since yellow and round are dominant,

Let G = yellow, g = green, W = round, w = wrinkled. Confirm on your own using a Punnett square! Dihybrid Cross: F1 cross Dihybrid Cross: Summary 9

3 3 1 Dihybrid Testcross: How to determine the genotype of an individual with 2 traits of dominant phenotype All yellow

Mixed All Round IV. Independent assortment Mendels 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. 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 V. Gamete Formation: Trihybrid cross What size of Punnett square needed for analysis? Trihybrid Cross - Phenotypes Forked-line Method

27:9:9:9:3:3:3:1 VI. Molecular Basis of Mendels 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. Theodor Boveri Hypothesis: Removal of chromosomes should result in some change to organism. Did this in sea urchin

embryos, saw abnormal embryos. 1862-1915 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

Mendels genes, and proposed that genes are carried on the chromosomes. 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. 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? 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 1s = 1/6 X 1/6 = 1/36 We used product law when calculating probabilities by the forked-line method. 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 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

Recently Viewed Presentations

  • Electron Orbitals  Electrons fill atomic orbitals in a

    Electron Orbitals Electrons fill atomic orbitals in a

    Diagonal Rule. Steps: Write the energy levels top to bottom. Write the orbitals in s, p, d, f order. Write the same number of orbitals as the energy level. Draw diagonal lines from the top right to the bottom left....
  • Learning Objective To understand Q3: explain, comment on and ...

    Learning Objective To understand Q3: explain, comment on and ...

    Learning Objective: Q3: explain, comment on and analyse how writers use language… George Graham Vest was a US lawyer. This speech was given in court while representing a man who sued another for the killing of his dog.
  • EVOLVING INCIDENT MANAGEMENT An Analysis of Organizational Models

    EVOLVING INCIDENT MANAGEMENT An Analysis of Organizational Models

    Temporary promotions during fire season. One type of team. Members still supervised by Agency Administrators. Talk about how IMTs could be organized so that the "core teams" were funded by emergency funds, but supervised at the unit level by a...
  • Cross-referencing

    Cross-referencing

    Function of cross-referencing Indicate variant spelling headword where definition appears Indication that further information is to be found at headword named: Antonym given for consultation cross-reference to the noun sense of the headword static small capitals used to indicate headword...
  • Physical Science Big Idea 10: Forms of Energy

    Physical Science Big Idea 10: Forms of Energy

    Extension: If available, play Bill Nye's Classroom Edition DVD LIGHT AND COLOR. Go to Watch Program Section: Science Standards Clips, Clip 3 Light can be absorbed and changed to heat. ... her body will eventually break down the peach through...
  • Presentation title

    Presentation title

    Galapagos potentiator and correctors Katja Conrath, PhD Therapeutic Area Head UK CF Conference, September 2016, Nottingham 7 September 2016 CF Program aim Develop a combination therapy CFTR Combination Strategy Potentiator Correctors Two correctors with complementary MoAs GLPG1837 MSD1 MSD2 R...
  • Native American Exhibit - TeacherTube

    Native American Exhibit - TeacherTube

    It is a memorial to the Choctaw Native American soldiers in the Civil War. Meaning: 6 white crosses= high ideals ... Pacific NorthwestNative American Masks. ... Mask. Way to go! This exhibit was brought to you from the 4th graders...
  • 240 Chem Aromatic Compounds Chapter 6 1  The

    240 Chem Aromatic Compounds Chapter 6 1 The

    It contains a six-membered ring and three additional degrees of unsaturation. It is planar. All C—C bond lengths are equal. ... A hydrocarbon with a saturated chain and a benzene ring is named by choosing the larger structural unit as...