Download Bio1A Unit 2-3 Genetics Notes File

Mendelian Genetics
• Trait – description of organism
• Gene – a heritable unit that influences a trait
• Allele – Version of a gene
- Most basic form of an allele is as a functional version or a non-functional
version
• Genotype – Allele combination
• Phenotype – observed trait
Genotype
PP
Pp
pp
Phenotype
=
=
=
Purple
Purple
White
Homozygous – the genotype has two alleles that are the same
Heterozygous – the two alleles are different
Dominance
Dominant – an allele that masks or suppresses the expression of its partner
Recessive – the allele that is masked or suppressed
1. List Alleles
2. List Genotypes
3. List Phenotypes
Pedigrees
4. Enter known genotypes
(don’t skip ahead)
5. Solve unknowns (if possible)
Look to parents & children
6. Punnett Square (or probability rules)
For predicting only
• what percentage of children
• what percentage of sons
• what percentage of daughters
• 2nd child
Do All genes separately until punnett
Basic  Autosomal = 2 alleles
Allels  A & a
Genotypes  AA Aa aa
Dominance = Dominant allele masks Recessive
•Dominant Trait  A = trait
•Recessive Trait  a = trait
Incomplete Dominance:
All genotypes have unique phenotype
Aa has diff phenotype from AA & Aa
Codominant – 2 alleles mask a third
IA and IB are don’t mask each other but
both mask i
Multi-allelic – more than 2 alleles
More than 2 alleles  IA, IB, i
Genotypes  IAIA IAIB IAi IBIB IBi ii
Sex-linked = 1 allele, 1Y (place holder)
• Recessive sex-linked
 XA = Wt Xa = trait Y = ∅ but men have
it
XAXA XAXa XaXa XAY XaY
• Dominant  XA = trait Xa = wt
Polygenic = More than 1 gene
•do each gene separately
Incomplete dominance
• the phenotype of F1 hybrids is somewhere between the phenotypes of the two parental varieties
Codominance
• two dominant alleles affect the phenotype in separate, distinguishable ways
Multiple-Allele Inheritance
•
•
•
•
Genes that exhibit more than two alternate alleles
ABO blood grouping is an example
Three alleles (IA, IB, i) determine the ABO blood type in humans
IA and IB are codominant (both are expressed if present), and i is
recessive
• Note: Rh+ & Rh- is a separate trait, so blood typing is also polygenic
Polygenic
• Trait is determined by multiple genes
• Each gene follows the rules of Mendelian Inheritance, and should be
examined separately
Sex-linked Traits
• For a recessive sex-linked trait to be expressed
– A female needs two copies of the allele
– A male needs only one copy of the allele
• Sex-linked recessive disorders are much more common in
males than in females
XBXB
XBXb
XbXb
XBY
XbY
=
=
=
=
=
normal female ♀
normal female ♀
colorblind female ♀
normal male ♂
colorblind male ♂
Epistasis
• In epistasis, a gene at one locus (location) alters the phenotypic expression of a gene at a
second locus
• For example, in mice and many other mammals, coat color depends on two genes
• One gene determines the pigment color (with alleles B for black and b for brown)
• The other gene (with alleles C for color and c for no color) determines whether the pigment will
be deposited in the hair
Hallmarks:
2 separate traits
Should give 4 distinct phenotypes
Instead there are only 3
Typical 9:3:3:1 ratio is now a 9:3:4 ratio
or
1:1:1:1 ration is now 1:1:2
indicating to phenotypes have merged into 1
X Inactivation in Female Mammals
• In mammalian females, one of the two X chromosomes in each cell is randomly inactivated
during embryonic development
• The inactive X condenses into a Barr body
• If a female is heterozygous for a particular gene located on the X chromosome, she will be a
mosaic for that character. Some cell will have on X chromosome, some cells will inactivate the
other X chromosome.
Genomic Imprinting
• For a few mammalian traits, the phenotype depends on which parent passed along the
alleles for those traits
• Such variation in phenotype is called genomic imprinting
• Genomic imprinting involves the silencing of certain genes that are “stamped” with an
imprint during gamete production
Inheritance of Organelle Genes
• Extranuclear genes (or cytoplasmic genes) are genes found in organelles in the cytoplasm
• Mitochondria, chloroplasts, and other plant plastids carry small circular DNA molecules
• Extranuclear genes are inherited maternally because the zygote’s cytoplasm comes from the egg
Environmental Influence on Gene Expression
Probability
Probability that event “A” occurs (PA) is the number of ways A can occur (NA) divided by the
total number of outcomes (NT). PA = NA / NT
• PA can be from 0 (never happens) to 1 (always)
• In a coin toss, Pheads (probability of getting heads)
= number of ways to get heads (1) divided by total number of outcomes (2) = ½.
• Similarly, Segregation in a heterozygous (A/a) plant:
Probability of getting a gamete with “a” from an “Aa” parent =
number of ways to get a (1)
divided by total number of possible (A or a) outcomes (2) = ½
Probability =
Ways “x” can happen
Total # possibilities
Example: chance of getting little “t” gamete from:
Ways “t” happens 
Total possible 
TT
Tt
Tt
0
2
1
2
2
2
The Multiplication Rule (The “&” Rule)
the probability that two or more independent events will occur together is the product of
their individual probabilities
• Probability in an F1 monohybrid cross can be determined using the multiplication rule
Prob of getting tt child?
Logic context
In order to get a tt.
• Mother must give a t
• AND
• Father must give a t
•Probability of getting “t” from mother
is the number of ways it can happen
•Out of (over) the total number
of possible outcomes
♀
♂
Tt x Tt
Pt
From Tt
t happens 1
From Tt
There are 2 possible
outcomes T or t
x
Pt
Pt♀ = ½
Pt♂ = ½
½x½=¼
The Rule of Addition (The “OR” Rule)
• The rule of addition states that the probability that any (either) one of two or more exclusive
events will occur is calculated by adding together their individual probabilities
• The rule of addition can be used to figure out the probability that an F 2 plant from a monohybrid
cross will be heterozygous rather than homozygous
Prob of getting Tt child?
Logic context
In order to get a Tt
1. Mother must give a T
AND
Father must give a t
♀ Tt
x Tt
½x½ = ¼
PT♀
Pt♂
x
+
OR
2. Mother must give a t
AND
Father must give a T
♂
Pt♀
+
PT♂
x
½x½ = ¼
2/4
Probability Changes with changes in information, should be based on current knowledge
2 tongue curler are having a child, what is the
probability their child will not be a tongue curler?
TT = curler
Tt = curler
tt = non curler
TT or Tt
TT or Tt
?
Pt♂ · Pt♀ = ¼ x ¼
= 1/16
If they then have a non-curler child,
what is the probability their 2nd child will be a non-curler?
Tt
Tt
?
tt
Pt♂ · Pt♀ = ½ x ½
= 1/4