P number of ways that a given outcome can occur total number of possible outcomes Gene Interactions There are three main types of gene interactions 1 Multiple Alleles There are many traits that are controlled by more than two alleles for example eye colour in Drosophila is contro ID: 590211
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Slide1Slide2
PROBABILTY
P=
number of ways that a given outcome can occur
total number of possible outcomesSlide3
Gene Interactions
There are three main types of gene interactions:
1) Multiple Alleles
There are many traits that are controlled by more than two alleles for example, eye colour in Drosophila is controlled by four possible alleles.
The following phenotypes and dominance hierarchy is possible:
wild type (red) > apricot > honey > whiteSlide4
Multiple Alleles
Wild type- the most commonly seen trait
Mutant- non-wild-type traits Slide5
When dealing with multiple alleles, it is no longer necessary to use upper & lower case letters
both letters & upper case numbers
are used.
phenotype genotype
wild type E
1
Apricot E
2
honey E
3
white E
4
Predict the genotypes and phenotypes of the
F1
generation from the mating of wild type (E
1
E
4
) with apricot (E
2
E
3
).Slide6
Predict the genotypes and phenotypes of the F1 generation from the mating of wild type (E
1
E
4
) with apricot (E
2
E
3
).
E
1
E
2
E
2
E4
E1E3
E3E4
E1
E4
E2
E3
In the F
1
generation ½ are wild type; ¼ apricot; ¼ honeySlide7
Do Now
phenotype genotype
wild type E
1
Apricot E
2
honey E
3
white E4With the above information, make two heterozygous crosses what phenotypes do you get?Slide8
2) Incomplete Dominance
In some heterozygotes, both alleles of a pair are expressed in the phenotype. These alleles are said to be equally dominant. This lack of dominance is known as
incomplete dominance
.
eg.
P
1
Black
White
F
1
Grey
P2 Grey × Grey F2 Slide9Slide10
ww
WW
Ww
Ex. Snapdragons
An interaction between the alleles in the heterozygote shows an intermediate phenotype.Slide11
3)
Codominance
A form of incomplete dominance where two alleles are expressed in such a way that the effect of each is noticed separately in the phenotype.
Both parental phenotypes can be distinguished in the heterozygote offspring.Slide12
The expression of one allele does not mask the expression of another.
Ex. A red bull crossed with a white cow =
roan
calf
(calf has intermingled white & red hair)Slide13
Roan cattle and horses have both coloured and white hair.
Strawberry Roan
Blue RoanSlide14
The ABO blood group system is another example.
Inheritance of blood groups is determined by the gene “I” which has three different alleles, only two of which can occur at the locus at once. The alleles are responsible for producing antigens on the surface of the red blood cells, which determines the blood group.
Alleles A and B are co-dominant so that when they are both present, both A and B antigens are produced. Both A and B are dominant to O.Slide15
Blood Types (A, B, O, AB)
Allele I
A
– formation of blood factor A (antigen A)
Allele I
B
– formation of blood factor B (antigen B)
Allele I – no factors result
Genotype Blood Type
I
AIA or IAi AIBIB or IBi Bii OIAIB
ABSlide16
I
A
an I
B
are codominant and i is recessive.
Rh antigens are straight dominant vs. recessive.
Type A
Type B
Type AB
Type OSlide17
Blood types of North America
Blood types of North America with RH.Slide18
Example:
A mother with blood type A has a child with Blood type O. The father is blood type B. Indicate the genotypes of the parents.
Father can be
I
B
I
B
or I
B
i and mother can be I
AIA or IAi. To have a child that is type O (ii) both mother and father must contribute the i allele, therefore the father must be IBi and the mother is IAi Slide19
Case Study
Page 611: A Mystery of Blood TypesSlide20Slide21Slide22
Pedigrees
A chart or register showing a line of ancestors
Circles represent females, squares represent males, solid circles & squares represent those who have the trait being studied.
Horizontal lines between circles and squares represent
mating between male and female.
A vertical line joins parents and
children.
Pedigrees are often used to study sex-linked traits such as color-blindness and hemophiliaSlide23Slide24
PEDIGREE SUMMARY
This will help you determine a trait in a pedigree as with autosomal or X-linked:
Autosomal Dominant
must be in each generation
affected individuals transmit to minimum ½ of their offspring
males and females are equally affected
cannot have carriers, they will all be affected!Slide25
Autosomal Recessive
may skip a generation
affected offspring generally have normal (but heterozygous) parents
male and female are equally affectedSlide26
X-linked Dominant
very likely to be observed in each generation
females pass on to half of either sex
no transmission from father to son (only daughters)
X-linked Recessive
affect males more than females
no transmission from father to son
daughters of males are carriers
females pass onto ½ sons
affected females have affected fathers and carrier mothersSlide27Slide28
Pedigree analysisSlide29
Probability
Probability is the likelihood of an event happening.
Probability can be expressed by the following formula:
Probability =
# of chances for an event
# of possible combinationsSlide30
Therefore, when a coin is tossed, there are two possibilities – heads or tails.
What are the chances of getting heads?
What is the probability of two coins being tossed and getting heads?
Coin 1 –
Coin 2 –
1
head
/2
head/tail = 50 % chance
½ = 50 % chance
½ = 50 % chanceSlide31
The Rule of Independent Events –
Product of the probabilities of two separate events:
The Product Rule –
½ × ½ = ¼ therefore a 25% chance
The probability of two or more independent events occurring together is the product of the individual probabilities if each individual event occurs separately.
Chance has no memory. This means that previous events will not affect future events. Ex. If you tossed two heads in a row, the probability of tossing heads again will still be ½.Slide32
Dihybrid Crosses
Mendel also studied
two separate traits
with a single cross
by using the
same procedure
he had used for studying single traits.
Slide33
Mendel crossed a purebred yellow round pea with a purebred green wrinkled pea.
Pure breeding round = RR
Pure breeding wrinkled = rr
Pure breeding yellow = YY
Pure breeding green = yy
Genotype for the yellow, round parent is RRYY
Genotype for the green, wrinkled parent is rryySlide34
P
1
YYRR x yyrr
Purebred yellow round
× Purebred green wrinkled
F
1
Male gametes
Female gametes
The entire F1 generation has the genotype: YyRr and is phenotypically yellow round!Slide35
Now let’s cross the F1 generations with one another and see what we get…Slide36
9/16 Yellow Round
3/16 Yellow wrinkled
3/16
green
Round
1/16
green
wrinkled
The phenotypic ratio 9:3:3:1 is the ratio you will find in all heterozygous dihybrid crosses!Slide37
Example:
In summer squash, white fruit color is dominant “W” and yellow fruit color is recessive “w”. Another allele produces disc shaped fruit “S” while its recessive allele “s” yields sphere-shaped fruit. If a pure breeding white disc variety is crossed with a homozygous yellow sphere variety, the F1 are all white disc hybrids. If the F1 generation is allowed to mate, what would be the expected phenotypic ratio in the F2 generation?
Try this!Slide38
P
1
WWSS x wwss
F
1
All offspring will be WwSs
P
2
WwSs x WwSs
F
2Slide39
Rhesus Factor & Birth
P
1
: Female Rh- × Male Rh+
Baby is Rh+ because father is. Mother’s blood produces antibodies upon birth, (since blood mixes at birth). First baby is okay.
Second pregnancy- mom’s antibodies can now move across the placenta and cause baby’s RBC’s to clump (agglutinate) if second baby is also Rh+. This decreases oxygen delivery in the baby – “
blue baby
.”Slide40
What can be done?
Mom can be given an injection of a drug that inhibits antibody production immediately after delivery.
What happens if this is undetected?
Baby could be given a blood transfusion while in the womb. Fairly uncommon.Slide41
Blood Types & Rhesus Factor Question
R – dominant allele (Rh+)
r – recessive allele (Rh-)
Example: A woman homozygous for blood type A and heterozygous for the rhesus allele, Rh+, has a child with a man with type O blood who is Rh-. What is the probability that their child will have blood type A, Rh+?
There will be a 50% chance.Slide42
Do Now
Biology creatures have many strange traits. Blinking eyes “B” is dominant and burning eyes are
recessive
“b”.
Another allele produces
pleasant smelling pheromones “P”
while its recessive allele
“p”
yields
rotting fruit pheromones. If a pure breed blinking pleasant smelling biology creature is crossed with a heterozygous blinking pleasant smelling biology creature, what would be the expected phenotypic ratio? Slide43
Techniques used in order to produce the genotype or phenotype that you want:
Selective Breeding
: the crossing of desired traits from plants or animals to produce offspring with both characteristicsSlide44
Inbreeding:
the process by which breeding stock is drawn from a limited number of individuals possessing desirable phenotypes.
Hybridization:
Blending of desirable but different traits. Slide45
Gene Interaction
1) Many traits studied by Mendel were controlled by
one gene.
2) Some traits are regulated by more than one gene; many of your characteristics are determined by several pairs of independent genes –
polygenic
.
eg.) skin color, eye color and height, feather colour in parakeets
3) One trait controlled by more than one allele (
multiple alleles
).
Eg. Blood types, Drosophilia eye colourSlide46
4) Genes that interfere with the expression of other genes are called
epistatic
.
Example:
allele
B
produces a black coat color in dog
b
produces a brown coat color
A second gene, W prevents the formation of pigment w does not prevent colorSlide47
genotype
wwBb
would be black
genotype
WwBb
would appear white
W allele masks the effect of the B color gene
If wwBb is crossed with a WwBb, state the phenotypes produced.
8/16 = white
6/16 = black
2/16 = brownSlide48
5)
Complementary Interaction
occurs when two different genotypes interact to produce a phenotype that neither is capable of producing by itself.
Example:
Allele R produces a rose comb in chickens
Allele P (on a different chromosome) produces a pea comb.
R and P alleles both present = walnut comb
The absence of rose and pea alleles results in an individual with a single comb