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Genetics Tutorial

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Title: Genetics Tutorial


1
Genetics Tutorial
  • Introduction
  • Punnett Square 1 Trait
  • Punnett Square 2 Traits
  • Product Rule

2
In this tutorial, you will learn
  • Important terms used in genetics.
  • How to use a Punnett square to determine the
    outcome of a cross with
  • 1 trait or
  • 2 traits
  • How to use the product rule to determine the
    outcome of a cross with any number of traits.

Credits Figures and images by N. Wheat unless
otherwise noted. Funded by Title V-STEM grant
P031S090007.
3
Introduction
  • Information that will guide the development of an
    organism is contained in that organisms DNA.
    Every species has a characteristic number of DNA
    molecules called chromosomes.

4
Introduction
  • An individual receives one complete set of
    chromosomes from each parent, resulting in two
    complete sets. This is the diploid condition
    (2n).

5
Chromosomes
  • Chromosomes occur in pairs called homologous
    chromosomes.
  • One from each parent.

6
Genes are the functional unit of heredity
  • Chromosomes are made up of genes that code for
    traits.
  • A gene is found at a specific location or locus
    on a chromosome.

7
Genes Alleles
  • Different versions of genes are called alleles.
  • Purple flowers vs. white in pea plants
  • Gene flower color, allele white or purple

8
Genes Alleles
  • There can be any number of alleles for a given
    gene, although an individual can have only two
    alleles(one on each homologous chromosome).
  • A, B, O blood type in humans

9
Genes Alleles
  • Some traits are controlled by just one gene,
    others are influenced by many genes (polygenic).
  • Height in humans

10
A, B and O are _______ of the human blood type
__________.
Question 1
  • Genes, genes
  • Genes, alleles
  • Alleles, genes
  • Alleles, alleles

11
Sorry!
Question 1
  • That is incorrect.
  • Try again!

12
Congratulations!
Question 1
  • You are correct!

13
Homozygous Heterozygous
  • Since an individual has two sets of chromosomes,
    it will have two copies of each gene (one
    originally coming from each parent). These two
    copies may be the same allele, or they may be
    different.
  • Homozygous both alleles are the same.
  • Heterozygous two different alleles.

14
Which of the following represents the homozygous
condition?
Question 2
  • AA
  • Aa
  • aa
  • Both AA and aa

15
Sorry!
Question 2
  • That is incorrect be sure your answer is
    complete.
  • Try again!

16
Congratulations!
Question 2
  • You are correct!

17
Dominant Recessive
  • A trait is dominant if it is expressed in an
    individual with one or two copies of the allele
  • Purple flower color in peas P purple p white.
  • The dominant allele is represented by a capitol
    letter, recessive by the lower case letter.
  • PP homozygous dominant two copies of the
    dominant (purple) allele.
  • Pp heterozygous one purple allele, one white
    allele (flowers appear purple).

18
Dominant Recessive
  • The trait is said to be recessive if it is
    necessary for an individual to have two copies of
    the allele in order to express the trait.
  • pp two white flower alleles (homozygous).

19
Purple flower color in peas is dominant over
white. Which of the following pairs of alleles
would give purple flowers?
Question 3
  • PP
  • Pp
  • pp
  • Both PP and Pp

20
Sorry!
Question 3
  • That is incorrect.
  • Try again!

21
Congratulations!
Question 3
  • You are correct!

22
Now, which of the following pairs of alleles
would give white flowers?
Question 4
  • PP
  • Pp
  • pp
  • Both PP and Pp

23
Sorry!
Question 4
  • That is incorrect.
  • Try again!

24
Congratulations!
Question 4
  • You are correct!

25
Genotype
  • Genotype refers to the alleles that are actually
    present.
  • PP, Pp, pp in our flower color example.
  • The purple phenotype may have PP or Pp genotype.

Back to question 5
26
Phenotype
  • Phenotype refers to the visible or expressed
    characteristics of the trait.
  • What does it look like?
  • Purple or white for our flower color example.

Back to question 8
27
Heredity Passing on Traits
  • An individual can pass on genetic information to
    its offspring. In order to avoid doubling the
    number of chromosomes in each generation, cells
    must be created that carry only one set of
    chromosomes (haploid or 1n).
  • An individual can pass along either of the two
    alleles it carries for a trait, but not both.
  • A Pp individual can pass on either P or p.
  • These haploid cells (eggs or sperm) are formed
    during meiosis.

28
Meiosis
  • Meiosis is a type of cell division that occurs in
    two parts.
  • Before division starts, each chromosome makes a
    copy of itself. These identical copies (called
    sister chromatids) remain connected at the
    centromere.

29
Meiosis
  • In the first part of meiosis, homologous
    chromosomes are separated.
  • The cells are now haploid.
  • During the second part of meiosis, the sister
    chromatids are pulled apart.
  • This part is like regular cell division
    (mitosis).
  • The division of one diploid cell will result in 4
    haploid cells.
  • Each cell is different.

30
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31
Heredity
  • We can look at how traits are passed from one
    generation to another individually or two at a
    time using a Punnett square.

32
Heredity
  • For our example, we will use the ball python.
    There are many mutations that breeders want to
    incorporate into their animals.
  • Albino a simple recessive trait
  • Pinstripe a dominant pattern mutation

33
Punnett Square 1 Trait
  • First lets focus on the albino trait. It is
    recessive so
  • AA Aa individuals will have normal coloration.
  • aa individuals will be albino.

34
Punnett Square 1 Trait
  • In a monohybrid cross we will cross two animals
    that are heterozygous for albino.
  • Aa x Aa
  • We want to know, statistically, what kind of
    offspring to expect.
  • Each parent can donate only one allele for the
    albino gene.
  • A heterozygote (Aa) can donate either an A or an
    a not both.
  • An albino must receive an a from both parents.

35
Punnett Square 1 Trait
  • Place the alleles that may be donated by each
    parent across the top and along the sides.
  • Fill in the boxes
  • 1AA - normal
  • 2Aa normal, heterozygous for albino
  • 1aa - albino

Back to question 8
36
Which is the genotype?
Question 5
  • Normal or albino
  • AA, Aa, or aa
  • Both are considered to be the genotype.
  • Neither of these is the genotype.

37
Sorry!
Question 5
  • That is incorrect.
  • Find information on genotype.
  • Try again!

38
Congratulations!
Question 5
  • You are correct!

39
Lets try another example!
  • If we have a male that is heterozygous for albino
    and an albino female, what kind of offspring do
    we get?

X
40
What is the genotype of this pairing?
Question 6
  • Aa x Aa
  • AA x aa
  • Aa x aa
  • aa x aa

41
Sorry!
Question 6
  • That is incorrect.
  • Try again!

42
Congratulations!
Question 6
  • You are correct!

43
Which of these Punnett squares is correct?
Question 7
44
Sorry!
Question 7
  • That is incorrect.
  • Try again!

45
Congratulations!
Question 7
  • You are correct!

46
What are the phenotypes of the offspring?
Question 8
  • All normal appearing offspring
  • All albino offspring
  • 31 normal to albino
  • 22 normal to albino

47
Sorry!
Question 8
  • That is incorrect.
  • Review Phenotype
  • Review Punnett squares
  • Try again!

48
Congratulations!
Question 8
  • You are correct!

49
Punnett Square 2 Traits
  • We can also use the Punnett square to track two
    traits at once.
  • Remember each gamete (egg or sperm) will contain
    one allele for each trait.
  • So, the possible combinations of alleles that we
    will place on our Punnett squares will always
    have one letter for each trait.

50
Punnett Square 2 Traits
  • In a dihybrid cross, both animals are
    heterozygous for two traits here, albino
    (recessive) and pinstripe (dominant).
  • AaPp x AaPp
  • The parents will have normal coloration (Aa) and
    they will be Pinstripes (Pp).

51
Punnett Square 2 Traits
  • Again, we want to know, statistically, what kind
    of offspring to expect.
  • Each parent (AaPp) will donate either an A or an
    a allele for the albino gene and either a P or a
    p allele for the pinstripe gene.
  • So every gamete will always contain ONE A(or a)
    and one P(or p).

52
FOIL
  • We can use the FOIL method from math to be sure
    that we have all of the possible combinations of
    alleles.
  • First, Outer, Inner, Last

53
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54
Punnett Square 2 Traits
  • Next, we fill in each square. By convention, we
    put the alleles for one gene together followed by
    the second
  • Aapp not Apap
  • Also, any dominant alleles are placed before
    recessives.
  • AaPp not aApP

55
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56
Genotypes from Dihybrid Cross
  • The Punnett square gives us the genotypes that
    result from the cross.

57
Phenotypes from Dihybrid Cross
  • The phenotypes would be
  • 9 Pinstripe (A_P_)
  • 3 Normal (A_pp)
  • 3 Albino pinstripe (aaP_)
  • 1 Albino (aapp)
  • Albino is a recessive trait, while pinstripe is a
    dominant trait.
  • The 9331 phenotypic ratio is characteristic of
    a dihybrid cross.

58
Which of the following Punnett squares is correct
for this cross AaPp x aapp
Question 9
59
Sorry!
Question 9
  • That is incorrect.
  • Try again!

60
Congratulations!
Question 9
  • You are correct!

61
What is the ratio of phenotypes that would result
from the cross? AaPp x aapp
Question 10
  • 9 normal 3 albino 3 pinstripe 1albino
    pinstripe
  • 4 pinstripe 4 normal 4 albino pinstripe 4
    albino
  • All albino pinstripe
  • AaPp 4 Aapp 4 aaPp 4 aapp

62
Sorry!
Question 10
  • That is incorrect.
  • Try again!

63
Congratulations!
Question 10
  • You are correct!

64
The Product Rule
  • Punnett squares are very useful for tracking one
    or two traits, but they can become unwieldy when
    looking at more than two traits.
  • The product rule is a simple way to determine the
    likelihood of getting a particular result from
    any cross, regardless of the number of traits
    involved.

65
The Product Rule
  • To use the product rule, we determine the
    likelihood of getting each trait individually,
    then multiply those probabilities together.
  • Well use our dihybrid cross example to start
    with.
  • AaPp x AaPp

66
The Product Rule
  • We need to look at the traits separately
  • Aa x Aa
  • There would be a 1 in 4 chance of hatching an
    albino from this cross.
  • Pp x Pp
  • Pinstripe is dominant so ¾ of the offspring will
    be pinstripes.
  • What is the chance of getting an albino pinstripe
    from this cross?
  • ¼ x ¾ 3/16
  • This is the same result that we got using the
    Punnett square.

67
The Product Rule
  • We can look at as many traits as we want using
    the product rule.
  • Say we are interested in combining these 4
    traits
  • Pinstripe (dominant) (PP, Pp, pp)
  • Albino (recessive) (AA, Aa, aa)
  • Piebald (recessive) (BB, Bb, bb)
  • Hypo (recessive) (HH, Hh, hh)

68
The Product Rule
  • The parents have the following genotypes
  • AaPpBbhh x aappBbHh
  • Calculate probability of getting individual
    traits
  • Albino Aa x aa ½
  • Pinstripe Pp x pp ½
  • Piebald Bb x Bb ¼
  • Hypo hh x Hh ½
  • ½ x ½ x ¼ x ½ 1/32 chance of getting an
    animal that shows all 4 traits from this pairing.

69
What is the probability of getting a hypo albino
piebald animal from these parents AaBbhh x Aabbhh
Question 11
  • Remember, albino, piebald, and hypo are all
    recessive traits.
  • ½
  • ¼
  • 1/8
  • 1/16

70
Sorry!
Question 11
  • That is incorrect.
  • Try again!

71
Congratulations!
Question 11
  • You are correct!
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