GeneticsTerry 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 GroupsGenotypes A O

O AO AO

O AO AO

Blood GroupsGenotypes 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