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					Chapter 9 & 12
Fundamental of Genetics
& Human Genetics
What’s the story about these
         two boys?
What if I told you, these girls
   were fraternal twins?
 LE 14-12



Most genetic                                                       AaBbCc       AaBbCc
traits are
polygenic,
meaning, they
are traits
determined by                                    aabbcc   Aabbcc   AaBbcc AaBbCc AABbCc AABBCc AABBCC
multiple genes.
                                         20/64
They are not
simply dominant
or recessive
(like those that                         15/64
we’ll learn in
this unit).
                   Fraction of progeny




                                          6/64




                                          1/64
       Gregor Mendel
   Austrian Monk ~ late 1800’s
   Pollinated pea plants (2N=14) to look at inheritance
   Determined that (some) traits are controlled by
    single genes (alleles) – he called them
    “factors”….why?
   Remember, scientists were still unsure if
    chromosomes or protein were “genetic material”
Mendel’s Garden
    Mendel’s Work
    In cross-pollinating plants that either produce yellow or green pea seeds
     exclusively, Mendel found that the first offspring generation (f1) always
     has yellow seeds. However, the following generation (f2) consistently
     has a 3:1 ratio of yellow to green.




    *Examine this slide again – after we do a monohybrid cross*
    Mendel’s Laws
   According to Mendel’s principle of segregation, for any particular trait,
    the pair of alleles of each parent separate and only one allele passes from
    each parent on to an offspring. Which allele in a parent's pair of alleles is
    inherited is a matter of chance.




We now know that this segregation of alleles occurs during the process of
sex cell formation (i.e., meiosis ).
       Review: Alleles
    2 forms a gene located on homologous
    chromosomes




                   r       R
         Phenotype vs. Genotype
   Phenotype: (P:)
        Physical characteristics (physical appearance)
        E.g. Red Petals


   Genotype: (G:)
     Alleles Received
     E.g. RR                          r      R
        Homozygous vs. Heterozygous
   Homozygous-
       Two of the same alleles (on homologous
        chromosomes)
       e.g. RR or rr
   Heterozygous-
       Two different alleles (on homologous
        chromosomes)
       e.g. Rr
     Dominant vs. Recessive
   About 600 human traits exhibit simple
    dominance

   Dominant alleles “mask” Recessive alleles
    (even though they’re both expressed)
Dominant:
       Allele that is phenotypically expressed
       Usually represented by a capital letter
       e.g. T


   Recessive:
       Allele that is expressed, however, not phenotypically, a
        lower case letter
       e.g. t
Examples!
R = red petals            r = white petals

-List the Phenotypes:
  1)   RR= Red petals
  2)   Rr= Red Petals
  3)   rr= White petals
More Examples
R = red petals            r = white petals

   If an individual plant has white flowers, is it
    heterozygous or homozygous for that trait?
   Ans: homozygous recessive
   If I told you a red flower has two different
    alleles for color, what is it’s genotype?
   Ans: Rr = heterozygous
Widow’s Peak
   Recessive or Dominant?
   Let’s take a class poll!

   W = Widow’s peak
   w= No widow’s peak
Mickey Mouse
   Mickey mouse has a widow’s peak!
   What is Mickey Mouse’s genotype for
    Widow’s Peak?
   Ans: WW or Ww
Review: Where do we get our
alleles?
          Mom                   Dad


               r        R
                                       r    r

                        They can pass on…
                                                     Sperm
eggs
           r            R              r         r
                   Or                       Or
     Monohybrid Cross
        Rr x Rr (P1 Generation)

               Possible Gametes    Possible
                                   zygote
                R             r

           R   RR        Rr
Possible
Gametes
           r   Rr        rr
    R         r
                  RR –1/4 (25%)
R   RR   Rr
                  Rr – 2/4 (50%)

r   Rr            rr – 1/4 (25%)
         rr
Genotypic vs. Phenotypic Ratios
   Phenotype Ratio:
       Ratio of different physical traits
       Ex) Brown eyes ¼ or 25%
   Genotype Ratio:
       Ratio of the different possible alleles
       Ex) Tt = ¼ or 25%
    R = Roll Tongue
    r= can’t roll tongue

                     Genotype Ratio:
    R           r
                     RR –1/4 (25%)
R   RR     Rr
                    Rr – 2/4 (50%)

r   Rr              rr – 1/4 (25%)
           rr
                     Phenotype Ratio:

           Roll Tongue (75%)
           Can’t Roll Tongue (25%)
        Sample Problems
Daffy Duck is heterozygous for black feathers. Daisy
   Duck is homozygous for yellow feathers. Set up a
   punnett square and determine probabilities of their
   potential offspring.
(Both genotype and phenotype ratios!)


B = Black      b = yellow
Dihybrid Cross
   Looking at two different traits
   Ex) color and height
   “Di” means two
   “Hybrid” means combine
            How to make
           a dihybrid cross
R = Red     r = white
T = Tall     t = short
Example cross: RrTt x RrTt
**Make the gametes for each parent**
RrTt 
RT Rt rT rt (gametes) (possible flower sperm)
RrTt 
RT Rt rT rt (gametes) (possible flower eggs)
 RrTt                     x RrTt
                 RT   (gamete)   Rt   (gamete)   rT   (gamete)   rt   (gamete)



RT(gamete)       RRTT             RRTt            RrTT            RrTt

Rt    (gamete)
                 RRTt             RRtt            RrTt            Rrtt

rT   (gamete)
                 RrTT             RrTt            rrTT           rrTt

rt   (gamete)
                 RrTt             Rrtt            rrTt            rrtt
Phenotypic outcome of
F1 offspring (p:)

P:
     9/16 red tall
     3/16 red short
     3/16 white tall
     1/16 white short
    Sample test questions for
  the above cross: RrTt x RrTt

Q. What fraction of F1 offspring are homozygous
  dominant for both traits?
1/16

Q. What fraction of F1 offspring are heterozygous for
  both traits?
4/16
Perform the cross: RRTt x Rrtt
  First make the Gametes
  RT Rt RT Rt        x   Rt Rt rt rt
 RRTt                       x               Rrtt
                 RT   (gamete)   Rt   (gamete)   RT   (gamete)   Rt   (gamete)



Rt(gamete)       RRTt             RRtt           RRTt            RRtt

Rt    (gamete)
                 RRTt             RRtt           RRTt             RRtt

rt   (gamete)
                 RrTt             Rrtt           RrTt            Rrtt

rt   (gamete)
                 RrTt             Rrtt           RrTt            Rrtt
Phenotypic outcome
P:
8/16 Red Tall
8/16 Red short
Trihybrid Cross
for color (R= red), stripe (S = stripe), height (T = tall)


RrSsTt              x        RrSsTt
Gametes:
RST RSt RsT Rst rST rSt rsT rst                  x same
      RST      RSt      RsT      Rst      rST      rSt      rsT      rst
RST   RRSSTT RRSSTt     RRSsTT   RRSsTt   RrSSTT   RrSSTt   RrSsTT   RrSsTt


RSt   RRSSTt   RRSStt   RRSsTt   RRSstt   RrSSTt   RrSStt   RrSsTt   RrSstt


RsT   RRSsTT   RRSsTt   RRssTT   RRssTt   RrSsTT   RrSsTt   RrssTT   RrssTt



Rst   RRSsTt   RRSstt   RRssTt   RRsstt   RrSsTt   RrSstt   RrssTt   Rrsstt



rST   RrSSTT   RrSSTt   RrSsTT   RrSsTt   rrSSTT   rrSSTt   rrSsTT   rrSsTt


rSt   RrSSTt   RrSStt   RrSsTt   RrSstt   rrSSTt   rrSStt   rrSsTt   rrSstt


rsT   RrSsTT   RrSsTt   RrssTT   RrssTt   rrSsTT   rrSsTt   rrssTT   rrssTt


rst   RrSsTt   RrSstt   RrssTt   Rrsstt   rrSsTt   rrSstt   rrssTt   rrsstt
P: for the Tri-hybrid cross
red striped tall               27/64
red striped short              9/64
red no stripe tall             9/64
red no stripe short            3/64
white stripe tall              9/64
white stripe short             3/64
white no stripe tall           3/64
White no stripe short         1/64
The chart for all Punnet charts
    # traits         # boxes (possible zygotes)

    1 (monohybrid)     4

    2                  16

    3                  64

    4                  256

    5                  1024
        Probability Questions
Take out any US coin (one per lab group).
Flip the coin three times.
Did anyone result in the outcome: (T) (T) (H)?
What is the probability that when flipped, you will result
   in the following outcome: (T) (T) (H)?
Ans = 1/8
How? ½ * ½ * ½ = 1/8

What is the probability that when flipped, you will result
  in the following outcome: (H) (T) (H) (H) (T) (H)?
Ans = 1/64
Here’s how probability relates to Genetics!
1. When tossing two pennies, what is the probability of getting one
   head and one tail? When crossing two heterozygous (Pp) plants,
   what is the probability of getting a heterozygous plant?
2. How are the sides of the pennies similar to the alleles P and p in this
   example?
   Each side of the penny represents one possible outcome of a
   coin toss. P and p each represent one possible allele a parent
   can pass to an offspring. In both cases, the possible outcomes
   pair randomly in the same proportions.
      Sample test questions on
            probability
Refer to the cross: Rr x Rr
What is the probability of three tongue rolling F1
  offspring?
Ans: ¾ * ¾ * ¾ = 27/64

What is the probability of two heterozygous F1
 offspring?
Ans: ½ * ½ =   ¼
      Sample test questions on
            probability
Refer to the cross: RrTt x RrTt
What is the probability of two resulting F1 offspring
  that are heterozygous for both traits?
Ans: 4/16 * 4/16 = 16/256

What is the probability that two flowers, both
  homozygous recessive for both traits, will result?
Ans: 1/16 * 1/16 = 1/256
**For review only**
**For review only**
                                                   **For review only**




1. List the dominant phenotypes in this example. List the recessive
   phenotypes.
   Dominant: round and yellow; recessive: wrinkled and green
                                                    **For Review only




2. From this dihybrid cross, calculate the probability of getting a seed
   that is heterozygous for both shape and color.
           (RrYy)
                                  Female
                                  (affected)



                                Female (not affected)




Note: not all pedigrees use shaded, non shaded, or
half-shaded circles or squares
                            RR or Rr = Tongue Rolling
                            rr = non roller

Rr or rr        Rr or rr




Rr or rr                                            Rr
           RR      rr      Rr




                                               rr
 Rr        Rr              RR
Sex Linked Traits
   Any gene on an “X” chromosome
   Two (possible) inherited from maternal

   eg.   XR Xr
   One (possible) inherited from paternal

   eg.   XR Y
Sex Linked Traits


                    Phenotypic
                    outcome F1:
                    100% red eyed
                    flies
1. What are the
   possible F2
   genotypes for eye
   color of a female fruit
   fly? Of
   a male?
        Female; XRXR,
        XRXr, or XrXr;
        Male: XRY
        or XrY
Practice test questions for
Sex Linked Traits
Colorblindness is sex linked and recessive

Jenni and Jonny visits Dr. Apple and Jonny
  says “Both myself and Jenni’s dad are
  colorblind. What is the chance Jenni and I
  will have a colorblind child?”

Ans. ½ for both CB female or CB male
      ?                ?         Color Blindness (sex linked &
                                 recessive)
  XbY               XBXb         XBXB or XBXb or XBY = norm.
                                 XbXb or XbY = color blind




            ?                ?                          ?

 XBY      XBXb or   XbXb   XBXb                     XbY
          XbXb


  ?

XBXB or     XBXb           XbXb                  XBY
XBXb
Sample Pedigrees
    Sample Pedigrees




1. Are attached earlobes a dominant or a recessive trait? Explain.
   Attached earlobes are a recessive trait. All individuals with
   attached earlobes are homozygous for the recessive allele.
  Sample Pedigrees




2. What is the genotype of a colorblind male? What possible genotypes
   can his mother have? Explain.
  Colorblind male: XbY; mother: XBXb or XbXb. The mother must be
  a carrier of the recessive allele or colorblind herself.
Multiple Alleles:
Blood type                        Genotype

A                                 I AIA    or   I AIO

B                                 I B IB   or   I B IO

AB                                IAIB
O                                 IO IO
 Which allele(s) is/are dominant? Recessive? Codominant?
 IA and IB are dominant over IO
 IO is always recessive
 IA and IB are co dominant
Blood type antigens & antibodies
Blood Typing: “Jamie’s” Pedigree
Jamie is Type O
Jorell (bro) is Type AB
Mom is Type B
Dad is Type ?
Maternal Grandpop is Type O
Maternal Grandmom is Type ?
**Draw in notes**
    A                  B            Blood Types
                                    (Co-Dominant)
  IA IO             IB IO




   A        ?         O         ?                   B

  IA IO   IAIB or   IoIo    IA I O              IBIO
          IBIB or
          IBIO

 AB         AB                AB           O              B

IA IB       IA IB           IA I B       IOIO           IBIO
Genetic Disorders
Three methods

1) Inherited Gene(s) (faulty or missing): “run in families”
Examples
    Cystic Fibrosis (Autosomal recessive)
    Hemophilia (Sex linked & recessive)
    Colorblindness (Sex linked and recessive)
    Huntington’s Disease (Autosomal dominant)
    Sickle Cell Anemia (Incomplete Dominance)
    Tay Sachs (Autosomal recessive)
    Albinism (Autosomal recessive)
    PKU (autosomal recessive)
2) Nondisjunction Missing or extra chromosome caused
  by error in separating chromosomes during Meiosis.
“Chromosomal Disorders”

Examples
Down Syndrome (46 + extra #21) “Trisome 21”

Turners Syndrome (44 + X)

Klinefelter’s Syndrome (44 + XXY)
Down Syndrome Incidence
3) Random mutation error in copying DNA, caused
  by chance, radiation, carcinogens, etc..
Example Possible birth defects
          Possible types of cancer

4) Cancer genes (oncogenes): Normally, the cell
  cycle is controlled by genes for
 growth factor proteins (initiate cell cycle)

 tumor-suppressor proteins (inhibit uncontrolled cell

  growth)
A mutation in one of the above types of genes can
  be called an “oncogene” (cancer causing gene)
How Nondisjunction occurs
Transparency 12A-2
Transparency 12A-3
Transparency 12A-4
1. Describe how
   nondisjunction in
   meiosis I and in
   meiosis II differ.
   Meiosis I:
   homologous
   chromosomes fail to
   separate; meiosis II:
   sister chromatids
   fail to separate.
2. How is it possible for
   nondisjunction to
   occur and for some
   normal gametes to
   be produced?
   If nondisjunction
   occurs in meiosis II,
   two of the four
   resulting gametes
   will be normal
   haploid.
Barbara McClintock’s Jumping
genes are “Transposons”

 1940’s research – Nobel prize in 1983….
 45% of the human genome is composed of
   transposons and their defunct remnants.
 Transposons are genes which code for an
   enzyme which moves the transposon gene
   to somewhere else in the genome
 *can cause disruption*
Transparency 12B-2
1. How does the position of the transposon change as a result of this
   process?
   The transposon moves from its original location into the middle
   of another gene.
       Transparency 12B-5




2. What effect might this movement have on the other gene?
  It disrupts the other gene’s sequence, interfering with the
  production of any polypeptide that gene codes for.
      Chromosome mapping
Frequency of crossover exchange...
       exchange of chromatids pieces of a homologous pair
during synapsis at a chiasma...
                   is GREATER the FARTHER
                   apart 2 genes are

                     is proportional to
                   relative distance between 2
                   linked genes

                     Relative distance is
                   measured as... 1% crossover
                   frequency =
                    1 map unit of map distance
                            above CO
                   Freq = 8.5% + 8.5% = 17%
                   1% CrossOver Freq =
                    1 centiMorgan
Incomplete Dominance
Chromosomal Mutations
Chromosomal Mutations
Gene Mutations

				
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