BIOL 102 Lecture 9 Tetrad Analysis (review) Gene Interaction Part I Variation to Mendel Reading Material: pages 147-148; 211- 218; 226-232 Problems: Ch 4; 38-42 (similar problems were assigned for Lecture 8) Problems: Ch6; 18, 21, 23, 24, 40, 44
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BIOL 102
Lecture 9
Tetrad Analysis (review)
Gene Interaction Part I
Variation to Mendel
Reading Material: pages 147-148; 211-218; 226-232
Problems: Ch 4; 38-42 (similar problems were assigned for Lecture 8)Problems: Ch6; 18, 21, 23, 24, 40, 44
1. Determine if the segregation pattern for a locus is MI or MII type.
2. Calculate the genetic distance b/w locus and centromere using the formula:
m.u. b/w locus and centromere =1/2 (# MII asci) x 100% total # asci
3. If the problem involves multiple loci perform (1) and (2) for each locus.
Note: Map distance b/w centromere & locus is limited to 33 m.u.
Tetrad AnalysisTetrad Analysis
A) Mapping Distance b/w Gene Locus and Centromere
I. Independent assortment of two genes will result in 50% Parental Ditype (PD) and 50% Non-parental Ditype (NPD) or recombinant Ditype.
B) Mapping Distance b/w Two Gene Loci
P: ab x a+b+
II. Complete Linkage of two genes will result in only parental ditype (PD) ascus a+b+ a bX
a+b+a+b+a ba b
Parental ditype
a+b+
a b
a+b+
a b
a+b+a+b+
ab
ab
Complete linkage with NO crossover, we cannot map distance
P1P1
P2
P2
P2P1
B) Mapping Distance b/w Two Gene Loci
III. Linkage of two genes with crossover will result in number of PD >> number of NPD asci
There are four types of asci to distinguish:
i) Parental ditype (PD) - no crossover b/w loci
ii) Non-parental ditype (NPD) - all spores recombinant
iii) Tetratype (T) -1/2 spores are recombinant
iv) Parental ditype (PD) - with MII segregation, crossover
between one locus and centromere, but NO crossover b/w
loci.
B) Mapping Distance b/w Two Gene Loci
iii) Tetratype is the product of a single crossover or 2 different CO that involves 3- strands a+ b+ a bX
Tetratype (T)
a+ b+
a b+
a+ b
a b
a+b+a+b
ab+
ab
where 1/2 of the spores are non-parental or recombinant (R1 and R2)
a+ b+a+ b+a ba b
P1R1
R2
P2
P1 P2
B) Mapping Distance b/w Two Gene Loci
a+ b+a+ b+a b a b
CO 3-strands SCO 2-strands
iv) Parental ditype - with MII segregation - No crossover b/w loci. a+ b+ a bX
a+ b+a+ b+a ba b
Parental Ditype (PD)
a+ b+
a+ b+
a b
a b
a+b+ab
a+b+
ab
where both parental types are present in both halves of the ascus
P1P2
P1
P2
P1 P2
B) Mapping Distance b/w Two Gene Loci
Tetrad types with haploid progeny genotypes
with MII segregation
B) Mapping distance b/w two gene loci:
1. Determine the phenotype of each spore in an ascus and score ascus as PD, NPD or T type ascus
2. Calculate genetic distance b/w two loci using formula:
Distance m.u. = 1/2 (T + 6NPD) x 100 total # asci
C) Construction of a Linkage Map
Based on the calculated genetic distances between individual gene loci and centromere and the distance between the two gene loci, consider possible order of the gene loci and the centromere
a b
a b
Distance mapped using mapping b/w a gene locus and the centromere and mapping b/w two gene loci may not add precisely, but should be close
When mapping more than two gene loci, map two gene loci at a time relative to each other
Penetrance and ExpressivityPenetrance and Expressivity
• Penetrance - the frequency with which a genotype actually expresses a phenotype. A trait has low penetrance if few individuals with the gene(s) express it.
• Expressivity - the degree to which a trait is expressed in individuals having the gene(s). Severity of genetic diseases such as cleft palate in humans are examples.
Each circlerepresents oneindividual
Sex Limited Traits
• The expression of certain traits is only in one the sexes
• Traits encoded by genes on the autosomes but the character is expressed only in one sex
E.g.- milk production in cows
Genotypes: LL, Ll, and ll
Phenotypes:
Female milk no milk
Male no milk no milk
Sex Influenced Traits
• Expression of certain traits is influenced by the sex• Appears in both sexes but the frequency or degree
of the phenotypic expression is different between the sexes
• Traits encoded by genes on the autosomes. Examples
-Pattern of baldness (see next slide) -Cleft lip & palate- more frequent in males (2 males:1 female)
-Osteoporosis- more frequent in females (1 male:3 females)
Sex Influenced Traits
Pattern of Baldness
Genotypes: b+b+ bb b+b
Phenotypes:
Males nonbald bald bald
Females nonbald bald
nonbald
Effects of Environment on Phenotype
Phenotype results from the genotype interacting with the environment. Changes in the environment can alter the phenotype.
Example: Himalayan rabbit
Genotype Ch_
Raise animal at 30 C - get white extremities Raise animal at 25-30 C - get dark color on extremitiesRaise animal at < 25 C - get dark extremities and flank
Multiple AllelesMultiple Alleles
• Essentially all genes have not just two alleles, but many since there can be mutations at any of 1000 or more bases in a gene and result in many possible alleles
• Let N be the normal (wild-type) base and X a mutation.Three alleles are:– NNNNNNNNNNNXNNNNNNNNNNN– NNXNNNNNNNNNNNNNNNNNNNN– NNXNNNNNNNNXNNNNNNNNNNN
• Each diploid individual has either two copies of the same allele, or two different alleles of each gene
• Inheritance patterns with multiple alleles can be more complex because more than two alleles may be present in the two parents combined
Multiple Alleles Multiple Alleles ABO Blood Group InheritanceABO Blood Group Inheritance
ABO Blood Group InheritanceABO Blood Group Inheritance
• Alleles: IA, IB, i • A possible cross: IA IB x IA i
Will
be
solv
ed in
cla
ss
Multiple Alleles Results in Various GenotypesMultiple Alleles Results in Various Genotypes
Incomplete DominanceIncomplete Dominance
Example: Snapdragon flower color
Will
be
solv
ed in
cla
ss
Incomplete DominanceIncomplete Dominance
• The F1 has an intermediate phenotype to the parents
• The phenotypic ratio in the F2 is 1:2:1• The F2 ratio is 1:2:1 rather than 3:1 because the
phenotype of the heterozygote differs from both homozygotes
• Similar terms: partial dominance, semidominance• The heterozygote need not be precisely intermediate
- there is incomplete dominance whenever it is different and somewhat intermediate in phenotype
Summary of Dominance Relations for Measurement Traits
CodominanceCodominance
• Codominance usually involves a system in which the 2 alleles of a single gene have slightly different products, both of which appear in the expression of the phenotype. Usually molecular. Effects of both alleles are detected simultaneously.
CodominanceCodominanceCodominance: effects of both alleles are detected simultaneously
E.g.- MN blood group antigens
Alleles: LM and LN. Phenotype is "blood group"
P: blood group M x blood group N
LM LM LN LN
F1: blood group MN
LM LN• Both M and N antigens are present on red blood cells.• The products of the two alleles are (usually) expressed equally. • The A and B alleles of the ABO blood groups are also codominant.
Modified Mendelian RatiosModified Mendelian Ratios
• We have seen that variation in dominance and multiple alleles can cause deviations from the 3:1 phenotypic ratio observed by Mendel for single gene inheritance
• When two genes are segregating and influence the same trait, then interaction between the genes or gene products can cause deviations from expected dihybrid ratios
Codominant Loci Alter Genetic RatiosConsider a dihybrid cross involving a dominant genefor height (D tall > d short) and the codominant AB blood allele pair
Parents: Dd IAIB x Dd IAIB
Will
be
solv
ed in
cla
ss
Distinguishing One vs Two Gene InheritanceDistinguishing One vs Two Gene InheritanceOne gene: F2 progeny occur in multiples of 1/4.
– Possible ratios: 3:1, 1:2:1, 1:1:1:1
Two genes (with independent assortment): F2 progeny occur in multiples of 1/16.– Many possible ratios depending on the nature of
inheritance– Use Product Rule: Form ratios as product of ratios
for each gene, example: (3:1) x (3:1) = 9:3:3:1 (1:2:1) x (3:1) =3:6:3:1:2:1
– If two genes influencing the trait are heterozygous in either or both parents, then more complex ratios can result from the interaction between the genes (or gene products).
Two Genes: Complementary Gene Action
Products of 2 gene loci complement each other to produce a phenotype. One dominant allele at each locus is required for full expression of the phenotype.
A common mechanism is that the two genes specify enzymes that function in the same biochemical pathway.
This is also called duplicate recessive epistasis
Complementary Gene Action
Products of 2 gene loci complement each other to produce a phenotype. One dominant allele at each locus is required for full expression of the phenotype.
E.g.- flower color in sweet pea. Cross two different
true breeding lines with white flowers.
Parents: white #1 x white #2
F1: purple flowers
F1 x F1
F2: 9 purple : 7 white
Complementary Gene Action - Mechanism
Consider a biochemical pathway:
Enzyme A Enzyme B X Y P
• Enzymes A and B must both be present to get color.• Only dominant alleles specify active enzymes.• Complementary gene action occurs when dominant genes are present at both loci to produce the purple phenotype.
White White Purple pigment
Complementary Gene Action - Inheritance
Hypothesis: 2 loci, A > a and B > b
Parents: white #1 x white #2 aaBB AAbbF1: purple flowers AaBb
F1 x F1
F2: A_B_ A_bb aaB_ aabb 9 3 3 1 purple white white white
Note that only A_B_ has a functional gene at both loci9 purple flowers : 7 (3+3+1) white flowers
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