Download Genetics Tutorial

Genetics
Tutorial
Introduction
Punnett Square – 1 Trait
Punnett Square – 2 Traits
Product Rule
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.
Introduction
 Information
that will guide the
development of an organism is contained
in that organism’s DNA. Every species has
a characteristic number of DNA
molecules called chromosomes.
Introduction
 An
individual receives one complete set
of chromosomes from each parent,
resulting in two complete sets. This is the
diploid condition (2n).
Chromosomes
 Chromosomes
occur in pairs
called homologous
chromosomes.

One from each
parent.
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.
Genes & Alleles
 Different
alleles.

versions of genes are called
Purple flowers vs. white in pea plants
 Gene
= flower color, allele = white or purple
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
Genes & Alleles
 Some
traits are controlled by just one
gene, others are influenced by many
genes (polygenic).

Height in humans
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.
Question 2
Which of the following
represents the homozygous
condition?
 AA
 Aa
 aa
 Both
AA and aa
Question 2
Sorry!
 That
is incorrect – be sure your
answer is complete.
 Try again!
Question 2
Congratulations!
 You
are correct!
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).
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).
Question 3
Purple flower color in peas is
dominant over white. Which of
the following pairs of alleles
would give purple flowers?
 PP
 Pp
 pp
 Both
PP and Pp
Question 3
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 That
is incorrect.
 Try again!
Question 3
Congratulations!
 You
are correct!
Question 4
Now, which of the following
pairs of alleles would give
white flowers?
 PP
 Pp
 pp
 Both
PP and Pp
Question 4
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 That
is incorrect.
 Try again!
Question 4
Congratulations!
 You
are correct!
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
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
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
 These
Pp individual can pass on either P or p.
haploid cells (eggs or sperm) are
formed during meiosis.
Heredity
 We
can look at how traits are passed from
one generation to another individually or
two at a time using a Punnett square.
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
Punnett Square – 1 Trait
 First
let’s focus on
the albino trait. It is
recessive so:


AA & Aa
individuals will
have normal
coloration.
aa individuals will
be albino.
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.
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
Question 5
Which is the genotype?
 Normal
or albino
 AA, Aa, or aa
 Both are considered to be the
genotype.
 Neither of these is the
genotype.
Question 5
Sorry!
 That
is incorrect.
 Find information on genotype.
 Try again!
Question 5
Congratulations!
 You
are correct!
Let’s 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
Question 6
What is the genotype of this
pairing?
 Aa
x Aa
 AA x aa
 Aa x aa
 aa x aa
Question 6
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 That
is incorrect.
 Try again!
Question 6
Congratulations!
 You
are correct!
Question 7
Which of these Punnett
squares is correct?
Question 7
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 That
is incorrect.
 Try again!
Question 7
Congratulations!
 You
are correct!
Question 8
What are the phenotypes of
the offspring?
 All
normal appearing offspring
 All albino offspring
 3:1 normal to albino
 2:2 normal to albino
Question 8
Sorry!
 That
is incorrect.
 Review Phenotype
 Review Punnett squares
 Try again!
Question 8
Congratulations!
 You
are correct!
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.
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).
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).
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
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
Genotypes from Dihybrid Cross
 The
Punnett square
gives us the
genotypes that
result from the
cross.
Phenotypes from Dihybrid Cross
 The
be:




phenotypes would
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 9:3:3:1 phenotypic ratio is characteristic of
a dihybrid cross.
Question 9
Which of the following Punnett
squares is correct for this cross:
AaPp x aapp
Question 9
Sorry!
 That
is incorrect.
 Try again!
Question 9
Congratulations!
 You
are correct!
Question 10
What is the ratio of
phenotypes that would result
from the cross? AaPp x aapp
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
Question 10
Sorry!
 That
is incorrect.
 Try again!
Question 10
Congratulations!
 You
are correct!
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.
The Product Rule
 To
use the product rule, we determine the
likelihood of getting each trait individually,
then multiply those probabilities together.
 We’ll use our dihybrid cross example to
start with.

AaPp x AaPp
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.
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)
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.
Question 11
What is the probability of
getting a hypo albino piebald
animal from these parents:
AaBbhh x Aabbhh
 Remember,
albino, piebald, and hypo
are all recessive traits.




½
¼
1/8
1/16
Question 11
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 That
is incorrect.
 Try again!
Question 11
Congratulations!
 You
are correct!