BLG400 week 3 - pedigree analysis (copy)

a pedigree is a

diagram of a family’s relevant genetic features over multiple generations

what are pedigrees used to study

human genetics

* to infer single gene inheritance of human trait
* pedigree analysis reveals several patterns of single gene inheritance
* expected 3:1 and 1:1 rations seen in mendelian law of equal segregation may not apply due to small progeny sample size
* many human diseases are caused by mutations in single genes

simple trio

a family group of a mother, a father, and their child

autosomal recessive disorders

* people with disorder are of genotype a/a
* examples: PKU, albinism, cystic fibrosis
* most telling sign is when two unaffected parents produce affected child
* parents are heterozygous while child is homozygous recessive

effect of inbreeding on autosomal recessive inheritance

inbreeding increases chance of homozygous recessive progeny

how can we recognize recessive traits in a pedigree?

* affected individuals can result from unaffected carriers, particularly as a result of inbreeding
* all the children of two affected parents should be affected
* rare recessive traits show a horizontal pattern of inheritance
* several affected individuals in one generation but non before or after (those guys are carriers)
* recessive autosomal traits appear equally in both sexes
* pedigree looks rather bare


1. if the condition is rare, most people do not carry the abnormal allele
2. most of those people who do carry the abnormal allele are heterozygous for it rather than homozygous

autosomal dominant disorders

* those with disorder are of genotype A/a or A/A
* examples: pseudoachondroplasia, Huntington disease, polydactyly, piebald spotting, etc.
* heterozygotes are most common
* 1:1 mendelian ratio is expected when sample size is large

how to recognize dominant traits is pedigrees?

* affected children always have at least one affected parent
* vertical pattern of inheritance
* acquiring allele from an ancestor
* two affected parents can produce unaffected progeny
* dominant autosomal traits appear equally in both sexes
* phenotype appears in every generation because abnormal allele carried by a person must have come from a parent in the preceding generation

polymorphism

coexistence of two or more common phenotypes of a character/trait

* most natural populations show polymorphisms

morph

the alternative phenotype

* often inherited as alleles of a single autosomal gene

dimorphism

simplest polymorphism with just two morphs

* i.e. brown versus blue eyes, pigmented versus blond hair, sticky versus dry earwax

autosomal polymorphism pedigrees

* pedigrees for polymorphisms are diff from rare disorders as polymorphisms are common by definition
* i.e. ability to either taste the chemical PTC (TT. Tt) or not (tt)
* most outsiders marrying in are assumed to be heterzygous (carry at least one copy of recessive allele as it is common)
* unlike rare disorders, where outsiders are homozygous normal
* even recessive traits may show a vertical pattern of inheritance if the trait is extremely common in population

pedigrees of x-linked recessive disorders

* more males than females show rare trait under study
* female can inherit phenotype if both mother and father carry allele whereas male can inherit phenotype if only mother carries allele
* mutation never passes from affected father to son
* son gets Y chromosome from male and X chromosome from female
* no children of affected male show phenotype but all daughters of affected male are carriers
* normal female of unknown genotype is assumed to be homozygous unless stated otherwise

x-linked recessive: explain how you know with certainty that II-4 is a heterozygous carrier of the recessive c allele

* because all daughters of affected male are carriers
* they get affected x chromosome from father and unaffected x chromosome from mother

pedigrees of x-linked dominant disorders

* affected males pass condition to all daughters and none of their sons
* affected heterozygous females married to unaffected males pass trait to half their sons and daughters
* uncommon mode of inheritance

y-linked inheritance

* trait only observed in males (male-to-male transmission)
* all male descendants of affected individual will exhibit trait
* usually a mutation on SRY region

step 1 of calculating risks in pedigree analysis: Tay-Sachs disease

newly married husband and wife find out that each had an uncle with Tay-Sachs disease, a severe autosomal recessive disease, probability of the couple’s first child having Tay-Sachs can be calculated as follows

step one:

* husband’s grandparents must have both been T/t because they produced a t/t child (the uncle)
* husband’s father could be T/T or T/t, the relative probabilities of these genotypes must be 1/4 and 1/2, respectively
* bcs expected progeny ratio in a monohybrid cross is 1/4 TT, 1/2 Tt, and 1/4 tt
* 2/3 probability that the father of husband is a heterozygote as probability of him being homozygous recessive is disregarded

step 2 of calculating risks in pedigree analysis: Tay-Sachs disease

* husband’s mother is assumed to be T/T, because she married into family and disease alleles are generally rare
* if the father is T/t, then mating with mother was a cross T/t × T/T
* expected proportions in the progeny (including husband) are 1/2 TT, 1/2 Tt

product rule and step 3 of calculating risks in pedigree analysis: Tay-Sachs disease

* overall probability of the husband being a heterozygote must be calculated __**product rule**__
* %%__probability of two independent events both occurring is the product of their individual probabilities__%%
* probability of the husband being a heterozygote:
* probability of his father’s being a heterozygote = 2/3
* probability of his father having a heterozygous son = 1/2
* 2/3 × 1/2 = 1/3

step 4 of calculating risks in pedigree analysis: Tay-Sachs disease

* probability of the wife’s being heterozygous is also 1/3 in this case
* if both mother and father are T/t, mating would be a standard monohybrid cross and probability of their having a t/t child is 1/4
* **THIS IS NOT THE FINAL ANSWER**

step 5 of calculating risks in pedigree analysis: Tay-Sachs disease

* probability of mother being Tt is 1/3
* probability of father being Tt is 1/3
* probability of their child being tt is 1/4
* **all of which are independent events, so overall probability must be calculated**
* product rule:
* 1/3 x 1/3 x 1/4 = 1/36
* 1 in 36 chance of them having a child with Tay-Sachs disease