Genetics BIOL 3250
Cahoon – Lecture Topic 3


Extensions of Mendelian Genetics
Patterns of inheritance do not always follo...
Genetics BIOL 3250
Cahoon – Lecture Topic 3


Incomplete Penetrance
Phenotype does not match genotype




Multiple Alleles...
Genetics BIOL 3250
Cahoon – Lecture Topic 3

Genotype                Phenotype          Frequency Within Population
I AI A...
Genetics BIOL 3250
Cahoon – Lecture Topic 3

       If a fetus inherits the Rh-pos trait from the father and fetal blood e...
Genetics BIOL 3250
Cahoon – Lecture Topic 3




It doesn’t matter what blood type genes you have if you can’t produce the ...
Genetics BIOL 3250
Cahoon – Lecture Topic 3

Example 2: Squash color

Example: Fruit color in summer squash
A_ __ = white
...
Genetics BIOL 3250
Cahoon – Lecture Topic 3

Albino + gene product B = black + gene product A = agouti

Example 4: Summer ...
Upcoming SlideShare
Loading in...5
×

Extensions of Mendelian Genetics

824

Published on

0 Comments
1 Like
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
824
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
7
Comments
0
Likes
1
Embeds 0
No embeds

No notes for slide

Extensions of Mendelian Genetics

  1. 1. Genetics BIOL 3250 Cahoon – Lecture Topic 3 Extensions of Mendelian Genetics Patterns of inheritance do not always follow Mendelian ratios. How can allele differences result in dominant vs. recessive phenotypes? Incomplete or Partial Dominance A cross between parents with different traits may produce offspring with an intermediate phenotype Four-o’clock plant flower color F0 F1 F2 From a genetic point of view (the nomenclature reflects this) neither allele is fully dominant and phenotypic and genotypic ratios are equal From a biochemical point of view we can say that the dominant gene is responsible for the production of the red pigment. One R1 (red) allele makes half the amount of pigment that two R1 alleles would make. Human example - Tay-Sachs disease freq. of carriers = 1/110 general pop.; 1/52 Irish descent; 1/30 Ashkenazi descent This neurological degeneration is caused by a lack of an enzyme needed to degrade membrane lipids. The result is a buildup of new lipid material in neurological tissue. The dominant allele produces a functional enzyme called hexosaminidase A (hex A) The recessive allele produces a non-functioning enzyme A heterozygous carrier is phenotypically normal but only half the enzyme activity is detectable 1
  2. 2. Genetics BIOL 3250 Cahoon – Lecture Topic 3 Incomplete Penetrance Phenotype does not match genotype Multiple Alleles – three or more Among members of an entire species many alternative forms of a gene may occur. Generally multiple alleles have to be studied at a population level since individuals can only carry two. Example 1: Rabbit Coat Color Example 2: Human blood types ABO blood group – three alleles which produce antigens on the surface of red blood cells Mix blood with antiserum containing type A or type B antibodies if A or B is present clumping (agglutination) will occur Allele IO = recessive allele = O blood type Allele IA = dominant allele = A blood type Allele IB = dominant allele = B blood type IA and IB are co-dominant = AB blood type 2
  3. 3. Genetics BIOL 3250 Cahoon – Lecture Topic 3 Genotype Phenotype Frequency Within Population I AI A I AI O 40% IBIB IBIO 10% I AI B 4% I OI O 46% The A and B genes code for glycosyl transferase enzymes which attach a monosaccharide subunit to a polysaccharide on the surface of red blood cells called the H-substance Rh antigens About 85% of the population carries the rhesus (Rh+) red blood cell antigen 15% do not (Rh-). The Rh antigen is a non-glycosylated cell surface protein. Rh is believed to be caused by multiple alleles across three gene sets. Responsible for HDN (hemolytic disease of the newborn) (10% of pregnancies are at risk but only ~0.5% of these are affected) This can occur when a Rh- woman and Rh+ man have children. 3
  4. 4. Genetics BIOL 3250 Cahoon – Lecture Topic 3 If a fetus inherits the Rh-pos trait from the father and fetal blood enters the mother’s bloodstream during birth, the mother will produce antibodies against the Rh-pos antigen. This makes the mother immunologically incompatible with future Rh+ fetuses. The mother’s antibodies which pass across the placenta will destroy the fetus’ red blood cells causing a form of anemia. Epistasis – Two genes are necessary to produce one trait Typical dihybrid F2 ratio Genotype Phenotype 9/16 3/16 3/16 1/16 Epistatic dihybrid F2 ratio Genotype Phenotype 9/16 3/16 3/16 1/16 Epistasis - The effect of one gene or gene pair masks or modifies the effect of another gene or gene pair. The phenotypes of the progeny from these crosses can be divided into 16ths because we are essentially dealing with a dihybrid cross of two individuals who are heterozygotes for both genes. Example: Bombay phenotype – cannot make the H-substance (hh), appears to be type O even with an A or B phenotype Exceedingly rare – most people are HH or Hh This was discovered when an Indian woman whose parents were AB and A and whose siblings were A, B, and AB was found to have an O blood type. 4
  5. 5. Genetics BIOL 3250 Cahoon – Lecture Topic 3 It doesn’t matter what blood type genes you have if you can’t produce the h-factor. The H-factor gene is epistatic to the AB genes. The AB genes are hypostatic to the h-factor gene. Consider the cross IAIO Hh x IAIO Hh Consider the cross IAIB Hh x IAIB Hh phenotypes genotypes Expected ratios 4/16 Type O with epistatic 3/16 Type A event 3/16 Type B 6/16 Type AB Figure out all the genotypes that could give the phenotypes listed above. 5
  6. 6. Genetics BIOL 3250 Cahoon – Lecture Topic 3 Example 2: Squash color Example: Fruit color in summer squash A_ __ = white aaB_ = yellow aabb = green Consider the cross AaBb x AaBb for summer squash color phenotypes genotypes Phenotypic outcome 12/16 white A_B_ A_bb 3/16 yellow aaB_ 1/16 green aabb Example 3: Purple corn kernel color Example 1: Purple corn kernel color A_B_ = purple aaB_ = white A_bb = white Consider the cross AaBb x AaBb for maize kernel color phenotypes genotypes Phenotypic outcome 9/16 purple A_B_ 7/16 white A_bb aaB_ aabb Colorless precursor + gene product A = colorless intermediate + gene product B = purple Example 2: Mouse coat color A_B_ agouti (brownish color) each strand of hair is an alternating black and yellow striped pattern. aaB_ black A_bb albino Consider the cross AaBb x AaBb for mouse coat color Phenotypic outcome 9/16 agouti 3/16 black 4/16 albino 6
  7. 7. Genetics BIOL 3250 Cahoon – Lecture Topic 3 Albino + gene product B = black + gene product A = agouti Example 4: Summer squash shape aabb = long (gooseneck) A_bb = sphere aaBb = sphere A_B_ = flat Consider the cross AaBb x AaBb for summer squash color Phenotypic outcome 9/16 disc 6/16 sphere 1/16 long Lethal Alleles Conditional Alleles 7

×