Terry Kotrla, MS, MT(ASCP)BB

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Genes
Chromosomes
Autosome
Sex chromosome
Locus
Alleles
Homozygous
Heterozygous
Antithetical

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Polymorphic
Syntenic
Cis and Trans
Linked
Crossing over
Phenotype
Genoytpe
Independent segragation

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Basic unit of inheritance.
hold the information to build and maintain an organism's cells and pass genetic traits to offspring.

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 organized structure of DNA and protein that is found in cells
46 total
23 homologous
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22 autosomes
1 sex chromosome – X or Y
XX female
XY male

 Location of a gene on a chromosome

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Homozygous – both alleles the same
Heterozygous – alleles are different

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Term used to describe allelic antigens and means opposite.
Kp a and Kp b are alleles.
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If an individual is heterozygous there will be one copy of each on “opposite” chromosomes.
Kp a is then said to be antithetical to antigen
Kp b

 Term used to describe having 2 or more alleles at a given locus.
 ABO system is an example
 Rh blood group system is highly polymorphic because of the greater number of alleles.

 Two genetic loci have been assigned to the same chromosome but still may be separated by a large enough distance in map units that genetic linkage has not been demonstrated.

 Genetic linkage occurs when particular genetic loci or alleles for genes are inherited jointly.
 Genetic loci on the same chromosome are physically close to one another and tend to stay together during meiosis, and are thus genetically linked

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Alleles at loci linked but sited at some distance from each other will often be separated by crossing over .
Crossing over happens at the first meiotic division of gametogenesis.
offspring that have different genetic make up from each other as well as different from either parent

 Alleles at loci which are carried on different chromosomes or at loci far apart on the same chromosome, and whose entry together into a sex cell is a matter of chance, are said to segregate independently .

 A phenotype is the assortment of antigens actually detectable on an individual's red cell.
 Genotypes cannot be determined with certainty and can only be accomplished through family studies.

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Example on left unless you knew parents type you would not know children’s genotype.
Example on right knowing children’s phenotype you can determine parents genotype.

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The mother in the first generation has the AB genes since her phenotype is AB.
In the mating for the second generation, the genotype for the father could either be BB or BO since his father's phenotype is unknown. It would appear the mother is AA since both her parents are A, but.....
Look at their children's blood types. What is the mother's genotype now since both children are B?

 Traits are the observed expressions of genes.

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Dominant – gene present trait will be expressed
Recessive – trait will not be expressed unless present in a double dose

 Codominant
 If the allele is inherited it will be expressed.
 Zygosity does not matter.

Genotypes A
O
O
AO
AO
O
AO
AO

Genotypes A
B
O
AB
AO
O
BO
OO

 Codominant traits are useful.
 Two types of exclusions
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Direct
Indirect

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Genetic marker present in child absent from mother and alleged father
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Child A
Mother O
 Father O
The A gene MUST have come from the “real” father as both parents are O.

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Genetic markers are absent from the child that should be transmitted by the alleged father
 Child Fy(a+b=) = Fy a /Fy a
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Mother Fy(a+b=) = Fy a /Fy a
Father Fy(a=b+) = Fy b /Fy b
Child should have inherited Fy b not there.
from dad but it is
Indirect because absence could be due to some rare
Fy genes that cause suppression of expression of Fy .
Dad may really be heterozygous Fy/fy and child inherited the fy gene making it appear homozygous.

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Direct exclusions much better.
DNA testing now the “gold standard”, very cheap and relatively fast.

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Important when attempting to find compatible blood.
Phenotype frequencies performed by testing a population and determining frequency of presence and absence of certain alleles.
Phenotype frequencies should be 100%
 Jk a+ = 77%
 Jk a= 23%
 23% of the population would be compatible for a patient with Jk a antibodies.

 Some patients have antibodies against
MULTIPLE antigens.
 Knowing the frequency of each antigen allows one to calculate the number of units which would need to be screened to find antigen negative blood.
 Performed by multiplying the percentages of each antigen negative allele.

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Must be able to perform calculation on next Exam .
The following frequencies are found:
Antigen Frequency of individuals negative for antigen c
K
Jk a
20% (0.20)
91% (0.91)
23% (0.23)
0.20 x 0.91 x 0.23 = 0.04 x 100 = 4 out of 100 units