1. Material from Thursday September 19th handout covered on Tuesday September 24th

2. Human Pedigrees: Polymorphic Traits
Polymorphic traits are alternative traits associated with alternative alleles at a single genetic locus
- example of dimorphism above is ability to taste the bitter compound, PTC (phenylthiocarbamide aka phenylthiourea or PTU)
- another commonly encountered polymorphism is blood type (ABO blood group locus)
Pedigree analysis:
Taster trait appears in every generation
Two taster parents (I-3 and I-4) and (II-1 and II-2) are unrelated
- common trait in the population: polymorphic trait
These two taster parents have non-taster children
- taster trait must be dominant; non-taster trait must be recessive
Male to male and female to female and male to female transmission
- not X or Y-linked: autosomal
Taster trait is most likely autosomal dominant
Polymorphic Loci:
Genetic loci can also be called polymorphic:
- different from a locus with one wild type and one mutant allele
- mutant alleles are typcially present at low frequencies, < 1%
- each alleles at polymorphic loci are present at ~ 2% or more
- examples of 3 allele loci that are present approximately equally = A, B, O blood group

3. Human Pedigrees: X-linked Dominant
Pedigree analysis:
Rare trait is present in each generation: rules out recessive
Transmission patterns include male to all female children; female to male children, female to some female; but no male to male: classic X-linked pattern
Trait is X-linked dominant
Example: hypophosphatemia (aka vitamin D-resistant rickets).
An abnormally decreased level of phosphates in the blood.
Progressive ankylosis (fusion) of spine and major joints
Spinal stenosis (narrowing)
Bowed legs
Molecular mechanisms of disease: not known...

4. Categories of inheritance
Autosomal recessive
e.g., PKU, Tay-Sachs, albinism
Autosomal dominant
e.g., Huntington disease
X-linked recessive
e.g., color-blindness, hemophilia, DMD
X-linked dominant
e.g., hypophosphatemia
Y-linked
Organelle
We will only consider these possibilities when analyzing pedigrees at this point in the course. We will exclude inbred families and consanguinous matings, which can cause pedigrees of autosomal recessive disorders to look similar t to X-linked recessive or autosomal dominant. Later, other possibilities will be covered, for example, when autosomal inheritance involves genes on the pseudoautosomal region of the X chromosome, or when X-chromosome inactivation reveals recessive X-linked traits in women.
Y-linked: male to male transmission only
Organelle – coming up…

5. Calculating probabilitiesAa Aa
2/3 Aa AA AA aa
2/3 Aa
2/3 x 1/2 Aa
2/3 x1/2 Aa
The trait appears to be autosomal recessive and is assumed to be rare.
The trait is assumed to be inherited from I-1 or I-2, who both must be heterozygotes.
Non-relatives of I-1 or I-2 are assumed to be homozygous for dominant allele.
The probability that each grandmother is heterozygous is 2/3
The probability that each parent of the offspring is heterozygous is 2/3 x ½.
If each is heterozygous, the probability of an afflicted child is ¼.
Therefore, using the product rule:
the probability that the child will be afflicted is 2/3 x ½ x 2/3 x ½ x ¼ = 1/36 aa
What is the probability that the offspring in question will have the trait?

6. Inheritance of Organelle Genes and Organelle-linked Traits
In humans:
2 rRNA
22 tRNA
13 proteins
Mitochondria and chloroplasts
small number of genes on circular chromosome
mostly inherited through maternal lineage via egg cytoplasm
Examples
white green variegation in plants
poky mutant in Neurospora
suspected examples in humans – we’ll cover this and not the two above

7. Inheritance of Traits due to Mitochondrial DNA Mutations
Maternal Inheritance:
- Affected females pass trait to all
- Affected males never pass on the trait

8. Material from Thursday September 24th’s handout covered Thursday September 24th

9. Chapter Overview
In meiosis, recombinant products with new combinations of parental alleles are generated by:
independent assortment (segregation) of alleles on nonhomologous chromosomes.
crossing-over in premeiotic S between nonsister homologs.
In Dihybrid meiosis, 50% recombinants indicates either that genes are on different chromosomes or that they are far apart on the same chromosome.
Recombination frequencies can be used to map gene loci to relative positions; such maps are linear.
Crossing-over involves formation of DNA heteroduplex.

10. Recombination:
A fundamental consequence of meiosis
independent assortment (independent segregation)
crossing-over between homologous chromatids
Yields haploid products with genotypes different from both of the haploid genotypes that originally formed the diploid meiocyte
Parental Vs Recombinant Progeny
Recombinant Progeny due to independent segregation o f unlinked genes or crossing over between linked genes
Independent segregation of genes on different chromosomes is a consequence of independent alignment of chromosomes in meiotic bivalents. Recombinant = 50% of progeny
Aa or Aa
Bb bB

12. Dihybrid Test Cross ¼ progeny proportions a b a/a ; b/b ¼ a ; b a B
Determines genotype of Dihybrid by crossing to homozygous recessive tester
Parents = pure breeding for two traits (genes on different chromosomes)
F1 = expected heterozygote for both genetic loci, dominant traits expressed
Test cross to homozygous recessive for both traits
Best way to study recombination is in a Dihybrid testcross
only Dihybrid produces recombinant genotypes
all homozygous recessive tester gametes alike
Typical 1:1:1:1 ratio a result of independent assortment in Dihybrid
Each genotype in progeny has unique phenotype
Ratio of phenotypes in progeny from test cross used to deduce genotype of parent: 1:1:1:1 indicates parent was heterozygous for both A/a and B/b
progeny proportions
a b
a/a ; b/b
¼ a ; b
a B
a/a ; B/b
¼ a ; B
A b
A/a ; b/b
¼ A ; b
A B
A/a ; B/b
¼ A ; B
progeny phenotypes
tester gametes
a ; b
F1 gametes
1:1:1:1 ratio

13. Dihybrid Self Cross Phenotypic Ratios using Product Rule
Cross between two A/a ; B/b dihybrids
recombination occurs in both members of cross
recombination frequency is 50%
Multiply probabilities of independent occurrences to obtain probability of joint occurrence
E.g. branched tree or grid methods
For mating A/a ; B/b  A/a ; B/b
Segregation at A, gives ¾ A/– and ¼ a/a in progeny
Segregation at B, gives ¾ B/– and ¼ b/b in progeny
Use ratio of phenotypes in progeny to deduce genotypes of parents: 9:3:3:1 ratio indicates parents are both heterozygous for A/a and Bb (R/r and Y/y above)
Phenotypic Ratios using Product Rule
1/16 a/a ; b/b
3/16 A/– ; b/b
¼ b/b
3/16 a/a ; B/–
9/16 A/– ; B/–
¾ B/–
¼ a/a
¾ A/–

14. Crossing Over creates recombinant progeny for genes on the same chromosome
Breakage and rejoining of homologous DNA double helices
No loss of genetic material, just formation of new chromatids
Detectable between nonsister chromatids at the same precise place (crossing over between sister chromatids occurs but does not cause change in genotype of progeny)
Visible in diplotene stage of meiosis as chiasmata (chiasma = 1 crossover)
Occurs between syntenic* and between linked loci on same chromosome
cis: recessive alleles on same homolog (AB/ab)
trans: recessive alleles on different homologs (Ab/aB)
(*Synteny: loci on the same chromosome, but not necessarily linked)

15. Cis – trans crossing-overb a B a b A B
cis
trans
meiotic crossing-over a B A b A B a b
Meiotic Products:
cis: trans:
AB; ab parental aB; Ab paraental
Ab; aB recombinant ab; AB recombinant
Parental chromatids are noncrossover products (with respect to A and B)
Recombinant chromatids are always products of crossing-over
All four genes (A, B, a and b) are present between both parental chromatids and between both recombinant chromatids (you must always account for all four meiotic products … nothing lost, nothing gained…)
Cis Dihybrid Crossing Over:
Parental (P) and recombinant (R) classes each have both alleles at each locus (reciprocal)
Each crossover meiosis yields two P chromosomes and two R chromosomes
For Linked Loci: CO does not occur in each meiocyte, frequency of recombinants (R) must be <50%
AB/ab  aB/Ab
Ab/aB  AB/ab
Drawing shows only chromatids engaged in crossing-over
Effect is to switch between cis and trans

16. Frequency of recombinant gametes is 0-50%, depending on frequency of meiocytes with crossing-over
Results in deviation from 1:1:1:1 in testcrosses when syntenic genes are linked
parental combination is most frequent
recombinant combination is rarest
Allows drawing of linkage maps based on recombination frequencies (RF)
Recombination frequency:
Experimentally determined from frequency of recombinant phenotypes in testcrosses
Roughly proportional to physical length of DNA between loci
Greater physical distance between two loci, greater chance of recombination by crossing-over
1% recombinants = 1 map unit (m.u.) 1 m.u. = 1 centiMorgan (cM)

17. Consequences of crossing-over
Genetic Analyses:
Frequency of recombinant gametes is 0-50%, depending on frequency of meiocytes with crossing-over (for syntenic genes)
Results in deviation from 1:1:1:1 in testcrosses
parental combination is most frequent
recombinant combination is rarest
Allows drawing of linkage maps based on recombination frequencies (RF)
Chromosome Segregation:
At least one CO per chromosome arm (in humans, usually only one CO per arm)
Required for equal segregation of homologs during meiosis I

18. HW/Study Problems topic 4 material covered through today’s lecture
Solved problems 1,2
Basic problems 1,2,3,4,5,6,17,18,19,26,