Mendelian Transmission Genetics

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• Mendelian (Transmission) Genetics
• Gregor Mendal (1822-1884) the founder of modern
  genetics- determined that discrete units of
  inheritance exist.
• Monohybrid cross
• Mate two parent strains-each exhibits one of the
  two contrasting forms of cell character under
  study.
• P1 X P2 Parent generation
•  F1 first filial generation
• F1 x F1
• F2 second filial generation
• Mendal performed all his significant genetic
  experiments with garden pea (Pisum sativum)-
  suitable organism for genetic experiments
• i)  Easy to grow
• ii) Self fertilizing in nature but could be
  easily hybridized artificially.

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• He first selected 7 traits to study in breeding
  experiments. Each trait had two easily
  distinguishable appearences (phenotypes)
• Phenotype the form of a trait that is expressed
  in a specific individual
• Results of Monohybrid cross
• round x wrinkled seed
  RR x rr
• F1 (all round)
  Rr
• F1 x F1
  Rr x Rr
• 5400 round 1800 wrinkled
  RR Rr Rr rr
• The wrinkled trait was hidden in the F1 3
  round 1 wrinkled
• The genotype expressing the hidden trait is
  recessive (r)
• The trait highlights in the F1 is dominant (R)
• Genotype ratio 12 1 phenotype ratio 31
• The composition of heredity characters (alleles)
  in an organism is its genotype

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• In a locus (gene site)- if both alleles are same-
  Homozygous RR rr if different
  heterozygous Rr
• Allele alternate forms of a gene
• Mendels first law of segregation- during gamete
  formation the paired unit factors segregate
  randomly so that each gamete receives one or the
  other with equal likelihood.
• Gamete formation-
• Rr x Rr
  
• R r R r
• Punnett square- generates the F2 ratio of the
  F1 x F1 cross
• R r phenotypic
  ratio genotypic ratio
• R RR Rr 3 round 1 wrinkled
  RR Rr rr
• r Rr rr ¾ round ¼ wrinkled
  1 2 1

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• To differentiate the genotypes of the 2 round
  phenotypes Test cross
• Cross with a homozygous recessive
• RR x rr Rr x rr
• Rr Rr rr
• all round ½ round ½ wrinkled

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• Dihybrid cross
• (Yellow wrinkled) seed x (green round) seed
• YY rr x yy RR
  Y, y, R, r alleles
• gametes Yr yR
• Yy Rr all yellow and
  round heterozygotes
• F2 generation resulted in
• 9 Yellow Round 3 yellow wrinkled 3 green
  round 1 green wrinkled (dihybrid ratio)
• Results are same for YYRR x yyrr
• Mendels second law of independent assortment
• When gametes are formed segregating pairs of unit
  factors assort independently of each other (fig)
• F2 generation phenotypes and genotypes - can
  determine in two ways
• Punnet square or branching method

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• Punnett square for a F2 dihybrid assume
• F1 is AaBb tall /short A/a
  brown/green seed coat B/b
• AABB
• AABb
• AaBB
  AaBb
• AaBb
• AABb
  AB Ab aB ab
• AAbb
  AB
• AaBb AaBb Ab
• Aabb
  aB
• AaBB
  ab
• AaBb
• aaBB
• aaBb
• AaBb
• Aabb
• aaBb
• aabb

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• genetic predictions can be made using
• Multiplication Rule ( product rule, and) if
  two events are independent, meaning one does not
  affect the other, then the probability that both
  will occur is the product of each probability ex
  probability of a family with 2 children having 2
  sons (one son AND one son) 1/2 x 1/2 1/4
• Addition Rule ( sum rule) "OR" if two outcomes
  are mutually exclusive, meaning that they cannot
  both occur at the same time, then the probability
  that either will occur is the sum of their
  probabilities
• probability that a woman who gives birth will
  have a child of either sex (a daughter OR a son)
  1/2 1/2 1

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• Calculating the probabilities
• From Mendals first law, for the gene Yy , half
  the gametes will contain Y allele, half will
  contain the other allele y
• Therefore p ½ for the occurence of either Y or
  y allele being present.
• What is the probability of a particular gamete
  type if independent assortment is occuring
• Events occuring independently of each other use
  the product rule
• Therefore p (RY) p(Ry) p(rY) p(ry) ½ x ½
  ¼

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• Branching method phenotypic ratio
• F1 yellow round x yellow round
• F2 of all offspring of all offspring
  combined probabilities
• ¾ round
  (¾)(3/4) 9/16 ye, ro
• ¾ yellow
• ¼
  wrinkled (3/4)(1/4) 3/16 ye, wr
• ¾ round
  (1/4)(3/4) 3/16 gr, ro
• ¼ green
• ¼
  wrinkled (1/4)(1/4) 1/16 gr, wr

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• Genotypic ratio
• ¼ YY
  1/16 RRYY
• ¼ RR ½ Yy
  2/16 RRYy
• ¼ yy
  1/16 RRyy
• ½ Rr ¼ YY
  2/16 RrYY
• ½ Yy
  4/16 RrYy
• ¼ yy
  2/16 Rryy
• ¼ rr ¼ YY
  1/16 rrYY
• ½ Yy
  2/16 rrYy
• ¼ yy
  1/16 rryy

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• Test cross
• YYRR x yyrr all yellow round
• YyRr x yyrr YR Yr
  yR yr
• yr YyRr
  Yyrr yyRr yyrr
• ye ro 1 ye wr 1 gr
  ro 1 gr wr 1
• Trihybrid cross Follow simple Mendelian
  inheritance for individual traits and apply the
  product law to calculate the probabilities in a
  branch diagram method. Phenotypic ratio would be
  as follows - 279993331
• Although Mendel published his findings in 1866
  the theory of units of inheritance did not agree
  with the continuous variation that was thought
  was a blend of parental phenotypes (Darwin and
  Wallace) . Discovery of chromosomes in 1879 by
  Flemming and Walter Sutton Theodor Boveri s

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• cytological work showed the behaviour of
  chromosomes during meiosis. Hugo de Vries, Karl
  Correns and Erich Tschermaks work on units of
  inheritance and principals of segregation
  reestablished Mendals findings.
• Due to independent assortment the number of
  genetically dissimilar gametes that an individual
  can produce is similar to 2n, if n is the haploid
  number. When n23, 223 or approx. 8 x 106
  possible gametes are formed, out of this only one
  is fertilized. The genetic variation due to
  independent assortment is an important process in
  evolution of organisms
• Assume that each trait is in a single chromosome
  
• n1 , of different gametes 2
• n2 , of ,, ,, 4
• n3 , of ,, ,, 8

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• Evaluating genetic data- Chi square analysis
• When sample sizes are small it is difficult to
  obtain predicted numbers of possible outcomes
  from crosses. The deviation from the predicted
  ratio in smaller sample sizes are due to chance,
  we test this using Chi square method
• Null hypothesis No real difference in predicted
  and observed ratios.
• ?2 S (o-e)2 /e d.f. no of classes-1
• Expected ratios
• Assume a dihybrid cross - expected is 9 3 31
• D.f. 4-1 3 (fig)
• If the probability p lt 0.05 -- significant -
  null hypothesis is rejected observed deviate
  from the expected ratios

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• Human pedigrees
• we cannot do experimental crosses with humans,
  so we must use pedigree analysis to understand
  genetics of human traits Conclusions from
  pedigree analysis whether allele is dominant or
  recessive (inheritance pattern).
• different patterns revealed in pedigree-
  dominant traits appear every generation,
  recessive traits may be present for a long time
  before being revealed.
• Therefore recessive traits are difficult to
  eliminate from a popn than a dominant trait.
• A family tree that shows the phenotype of the
  trait in question for each member of the family
  is named as a pedigree.
• Circles represent females, squares males, if sex
  is unknown a diamond

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• If a pedigree traces a single trait, the symbols
  are shaded if the trait is expressed
• Those who fail to express a recessive trait and
  are known to be hets, only the left half is
  shaded

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• Dominant Affected individuals can appear in
  every generation
• Autosomal Gene is on one of the autosomes (in
  humans, chromosomes 1-22). Male and female
  offspring equally likely to inherit trait.
• A trait that appears in successive generations is
  normally due to a dominant allele.

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• Trait can appear in offspring of related
  individuals
• Parents of affected children in pedigree shown
  are first cousins

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• 1) Assume that in a series of experiments,
  plants with round seeds were crossed to plants
  with wrinkled seeds and the following offspring
  were obtained 220 round 180 wrinkled. (a)
  What is the most probable genotype of each
  parent? (b) What genotypic and phenotypic ratios
  are expected? (c) Based on the information
  provided in part (b) above, what are the
  expected (theoretical) numbers of progeny of
  each phenotypic class?
• 2) How many possible types of gametes can be
  formed in F1 after a cross AAbbCCDDEEff x
  aabbccddeeFF ?

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• 3)Assume that in your garden some pea plants have
  solid leaves and others have striped leaves. You
  conduct a series of crosses and obtain the
  results given in the table.
  Cross Progeny
  
  Solid
  Striped (a) solid x striped 55
  60 (b) solid x solid 36
  0 (c) striped x striped 0
  65 (d) solid x solid 92
  30 (e) solid x striped 44
  0 Define gene symbols and
  give the possible genotypes of the parents of
  each cross

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• 4) Among dogs, short hair is dominant to long
  hair and dark coat color is dominant to white
  (albino) coat color. Assume that these two coat
  color traits are caused by independently
  segregating gene pairs. For each of the crosses
  given below, write the most probable genotype (or
  genotypes if more than one answer is possible)
  for the parents. It is important that you select
  a realistic symbol set and define each symbol
  below Parental Phenotypes
  Phenotypes of Offspring
  Short Long Short
  Long
  Dark Dark Albino Albino (a) dark,
  short x dark, long 26 24
  0 0 (b) albino, short x albino,
  short 0 0 102 33
  (c) dark, short x albino, short 16
  0 16 0 (d) dark, short x
  dark, short 175 67 61
  21 Assume that for cross (d) above you were
  interested in determining whether fur color
  follows a 31 ratio. Do a Chi-square test for
  these data (fur color in cross (d)).

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• 5) A man who had purple ears came to the
  attention of a human geneticist. The human
  geneticist did a pedigree analysis and made the
  following observations In this family, purple
  ears proved to be an inherited trait due to a
  single genetic locus. The man's mother and one
  sister also had purple ears, but his father, his
  brother, and two other sisters had normal ears.
  The man and his normal-eared wife had seven
  children, including four boys and three girls.
  Two girls and two boys had purple ears. The
  purple-ear trait is most probably
• Autosomal recessive
• Autosomal dominant
• Sex linked recessive
• Sex linked dominant
• Explain your result drawing a pedigree diagram.

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6) A pea plant exhibits the dominant phenotype
for two traitsits seed color is yellow and its
pods are swollen. A self-cross produces 178
progeny with the following phenotypes Yellow,
swollen 132 Yellow, pinched 46 Give the genotype
of the pea plant 7) In dogs, the tendency to bark
while trailing is due to a dominant gene while
silent trailing is recessive. Erect ears are
dominant to drooping ears. What kind of pups
would be expected from a heterozygous,
erect-eared barker mated to a droop-eared, silent
trailer? 8) In an effort to develop a
true-breeding strain of tall pea plants, a
geneticist student crossed two tall plants. All
of the 200 progeny (F1) of this cross were tall.
She then chose two plants from the F1 generation
and crossed them. All of the F2 generation was
tall. She crossed two
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• of the tall F2 plants and produced an F3
  generation of tall plants. However, in crossing
  two plants of the F3 generation, she got dwarf
  plants in the fourth generation. She would have
  dismissed it as a new mutation, but a fair
  proportion of the F4 plants were dwarf. Explain
  what happened and predict what proportion of the
  F4 generation were dwarf.