Transcript Genetics

Genetics
Terry Kotrla, MS, MT(ASCP)BB
Terminology










Genes
Chromosomes
Locus
Alleles
Homozygous
Heterozygous
Syntenic
Cis and Trans
Linked
Crossing over
Gene


Basic unit of
inheritance.
hold the information to
build and maintain an
organism's cells and
pass genetic traits to
offspring.
Chromosomes



organized structure of DNA and protein that is
found in cells
46 total
23 homologous




22 autosomes
1 sex chromosome – X or Y
XX female
XY male
Chromosomes
Locus

Location of a gene on a chromosome
Homozygous or Heterozygous


Homozygous – both alleles the same
Heterozygous – alleles are different
Syntenic

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.
Linked or Genetic Linkage


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
Crossing Over



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
Segregation

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.
Phenotype and Genotype


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.
Genotype versus Phenotype


In example on left unless you knew parents type you
would not know children’s genotype.
In example on right knowing children’s phenotype
you can determine parents genotype.
Genotype



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

Traits are the observed expressions of genes.
Dominant and Recessive


Dominant – gene present trait will be expressed
Recessive – trait will not be expressed unless present
in a double dose
Traits
Blood Group Antigens



Codominant
If the allele is inherited it will be expressed.
Zygosity does not matter.
Blood Groups
Genotypes
A
O
O
AO
AO
O
AO
AO
Blood Groups
Genotypes
A
O
B
AB
BO
O
AO
OO
Parentage Testing


Codominant traits are useful.
Two types of exclusions


Direct
Indirect
Direct Exclusion

Genetic marker present in child absent from
mother and alleged father




Child A
Mother O
Father O
The A gene MUST have come from the “real”
father as both parents are O.
Indirect Exclusion

Genetic markers are absent from the child that
should be transmitted by the alleged father





Child Fy(a+b=) = Fya/Fya
Mother Fy(a+b=) = Fya/Fya
Father Fy(a=b+) = Fyb/Fyb
Child should have inherited Fyb from dad but it is
not there.
Indirect because absence could be due to some rare
Fy genes that cause suppression of expression. Dad
may really be heterozygous Fy/fy and child inherited
the fy gene making it appear homozygous.
Parentage Testing


Direct exclusions much better.
DNA testing now the “gold standard”, very
cheap and relatively fast.
Population Genetics



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%



Jka+ = 77%
Jka- = 23%
23% of the population would be compatible for a patient
with Jka antibodies.
Population Genetics



Some patients have MULTIPLE antibodies.
Knowing the frequency of each allele allows
one to calculate the number of units which
would need to be screened to find compatible
blood.
Performed by multiplying the percentages of
each antigen negative allele.
Population Genetics

The following frequencies are found:
Antigen Frequency of individuals negative
c
20% (0.20)
K
91% (0.91)
Jka
23% (0.23)
0.20 s 0.91 x 0.23 = 0.04 x100 = 4 out of 100 units
Exam 2 Online