Extensions of Mendelism

Recall that Mendel took great care in the selection of his breeding stock for hybrid experimentation. He carefully selected traits useful to his experimentatal approach and thus was able to demonstrate pedictable ratios in different generations. It is hardly surprizing that inheritance patterns different from, but related to, his observations have been made repeatedly over the years.

Dominance relationships

Incomplete dominance

When the heterozygote's phenotype is unlike either of the two homozygotes the dominance/recessive relationship between the two alleles has been altered. In the snapdragon, flower color of the heterozygote is in-between that of the two parents. When true breeding white and red flower stocks are crossed, the F1 can be seen to pink - as if an artist mixed red and white paint. The F2 segregates red : pink : white in a 1 : 2 : 1 ratio. 1 : 2 : 1 is the expected F2 genotypic ratio from a monohybrid F1. Thus in incomplete dominance the phenotypic ratio mimics the ? ratio and is simply explained by assuming that the monohybrid a1a2 segregates according to the Law of Segregation.

F1gametes a1 a2
a1 a1a1 a1a2
a2 a1a2 a2a2


Codominance

When both homozygous phenotypes are observable in the heterozygote the phenotypic expression is said to be codominant. This differs from incomplete dominance where the heterozygote also had a phenotype different than either homozygote but the individual homozygous parental phenotypes were not observed.

Codominance is seen in the inheritance of ABO blood group alleles. Red blood cells have complex glycoprotein antigens on their surface that result, in part, from the expression of specific blood group alleles. Allele IA determines one type of antigen and allele IB determines another type of antigen.

Jock has blood type A, is of genotype IAIA, and has red blood cells with type A surface antigens. Maud is blood type B, is of genotype IBIB, and has red blood cells with type B surface antigens. Their children would have only one genotype, IAIB and have typeA and type ? antigens on the surface of their red blood cells. Thus Jock's typeA phenotype and Maud's typeB phenotype are both expressed in their children which are of AB phenotype. The children's red blood cells have both typeA glycoproteins and typeB glycoproteins on their surface. Offspring of IAIB by IAIB have A:AB:B blood types in a ? ratio.

P gametes

IA
IB IAIB

F1 gametes IA IB
IA IAIA IAIB
IB IAIB IBIB

Dominance Summary: fill in the blanks with one of complete, codominant, incomplete

Dominance P1 P2 F1 Dominance expression
complete a1a1 a2a2 a1a2 a1 is dominant 
? a1a1 a2a2 a1a2 a2 is dominant 
? a1a1 a2a2 a1a2 intermediate phenotype 
? a1a1 a2a2 a1a2 both phenotyes 

 


Lethal Alleles

At first thought it seems unlikely that an allele could result in a lethal phenotype! If the allele results in an organism's death, how does it continue to passed on to subsequent generations? There are several ways this can happen.

Recessive lethals

If the allele is recessive it is masked in the heterozygote and only expressed in the recessive homozygote. Wild-type AA agouti mice have black yellow-tipped hairs that give the animal a grey color. One allele of the agouti gene AY results in a yellowish color. P1 agouti crossed to P2 yellow results in a F1 yellow phenotype AY is a ? color allele. However, when two yellow F1s are crossed the F2 progeny segregate yellow : agouti in a 2:1 ratio. This deviation from the expected 3:1 F2 phenotypic ratio is explained by assuming a modification of the classical 1:2:1 F2 genotypic ratio. The AY is lethal when homozygous AYAY leaving only the heterozygous AAY yellow and AA agouti which segregate in a 2:1 ratio. AY has an ambivalent dominance acting as a ? color allele and a ? lethal allele. TSD is the recessive lethal allele for Tay Sachs disease, a progressive neurodegenerative disorder, which in the classic infantile form, is usually fatal by age 2 or 3 years. This disease in humans is expressed in early childhood rather that the zygotic death of AY.

 

F1 gametes AY A
AY

AYAY

AYA
A

AYA

AA

 

Conditional lethals

A conditional allele is one whose phenotype is expressed under one environmental condition, or life stage, but not another. Yeast geneticists have isolated vital genes by producing alleles with a temperature sensitive lethal phenotype. For example, cdc10 is a recessive mutant allele that will not undergo cell division at 36C but grows normally at 30C. This strategy is widely used and has allowed geneticists to identify genes involved in cell division. A brief description of the cdc10 phenotype is given at Saccharomyces Genome Database. Dominant lethal alleles can also be maintained in a population by being expressed after the normal reproductive reproductive period. Huntington disease in humans is inherited as a dominant allele HD so that a HDHD+has a 50% chance of passing this trait onto a child. However, since the disease is normally expressed in late life, the HDHD+carier could have unknowingly transferred Huntington disease to the next generation. Other examples include eye-cancer alleles such as RB for retinoblastoma. This bizzare, but deadly, allele behaves as a recessive cellular gene which is only expressed when a second cellular " mutation " occurs. Since secondary cellular "mutations" of RBRB+ heterozygotes to RBRB homozygotes almost always occur, RB is a cellular recessive that behaves as a dominant allele in pedigrees!

Allele   feature
AY mouse 1:2:1-> 2:1, recessive lethal ( ? color)
TSD human death in childhood
cdc10 yeast conditional - recessive lethal at 36C
RB human conditional - cellular ? ; pedigree dominant
HD human conditional -dominant, late onset expression

 

 


Multiple Alleles

The two allelic system used by Mendel was sufficient to show the fundamental units of heredity and the segregation of these units in successive generations. However all genes that are thoroughly investigated turn out to have numerous alleles, only some of which are expressed as a morphological trait similar to Mendel's wrinkled vs round seed traits.

Alleles show 3:1 F2 ratios with each other

Consider alleles of the agouti gene in mice.

P1 P2 F1 F2 segregating alleles
agouti yellow belly agouti 3 agouti : 1 yellow belly ? & ?
agouti black agouti 3 agouti : 1 black ? & ?
yellow belly black yellow belly 3 yellow belly : 1 black ? & ?
         

The three different F2 3:1 ratios show that in each cross two alleles of a coat color gene are segregating. This observation leads us to assign allelic genotypes to each of the three parental lines: agouti - AA; yellow belly - ay ay; black - ab ab. The F1s of the three crosses carry one allele from each of the parental lines, and the 3:1 segregation in the F2 clearly defines the allelic relationships. Only the first two crosses in the above table were required to establish the allelic series. Since A = ay and A = ab, then it follows that ay = ab. As expected the third cross confirmed allelism by the 3:1 segregation of yellow belly (ay_ ): black(abab).

The dominance hierarchy for the three alleles is ? > ? > ? The dominance hierarchy leads to the possibility of more than one genotype for a given phenotype. In this series of three alleles, agouti has ? possible genotypes, yellow belly has ? possible genotypes, and black has ? possible genotype.

ABO blood type allelic series

Successful blood transfusions depend on an understanding of the multiple allelism controlling antigens on the surface of red blood cells. The allelic relationships can be described as IA = IB > i. That is alleles IA and IB are codominant and are both dominant to the recessive allele i. Practice your understanding of the ABO multiple alleles by filling in the table below.

Genotypes

Phenotypes

(blood types)

IAIA

IA ?

?

IBIB

IB ?

?

IAIB AB
? ?

Practical application of the ABO alleles requires individuals to be typed to one of the four phenotypes A, B, AB, or O. Individuals carrying allele IA do not produce antibodies against IAand likewise IB individuals do not produce antibodies against IB. This is very sensible since antibodies against an antigen result in red blood cell agglutination and blockage of blood vessels. Oddly, anyone carrying only allele IA produces antibodies against IBand likewise anyone carrying only allele IB produces antibodies against IA. Consequently blood from a type A donor would agglutinate blood from type B and visa versa.

These considerations lead to the results outlined below. Fill in the permissible donors pattern (+) or (-) to the right of the table.

Recipient Blood Type

Serum Antibodies

Donor

A

Donor

B

Donor

AB

Donor

O

A antibodies against B + ? ? +
B antibodies against A ? + - ?
AB no A or B antibodies ? + + ?
O antibodies against A and B - ? ? +

Lucky those who are universal recipients, type ? , and everyone's friend is the universal donor - type  ?

Pleiotropy

Mendel's discription of the third trait he studied was:

"To the difference in the color of the seed-coat. This is either white, with which character white flowers are constantly correlated; or it is gray, gray-brown, leather-brown, with or without violet spotting, in which case the color of the standards is violet, that of the wings purple, and the stem in the axils of the leaves is of a reddish tint. The gray seed-coats become dark brown in boiling water."

 

He was aware that the gene that affected seed coat color also affected flower color. The recessive allele resulted in both white seed coat and white flowers whereas the dominant allele resulted in gray seed coat also resulted in violet flowers. He even noted a chemical component to the phenotype since boiling seed coats turned them a dark brown! When a gene affects more than one aspect of the phenotype it is said to be pleiotropic. This is very common since the action of any one gene is coordinated with numerous other genes. For example, a disease allele prevents the conversion of phenylalanine into tyrosine resulting in phenylketonuria. The phenotypic effect of phenylketonuria include mental retardation, organ damage and problems with pregnancy.

Expressivity

The dominant allele for seed coat had a variable phenotypic expression. It was not always solid gray but could also be gray-brown, leather-brown and any of these three colors came with or without violet spotting. Expessivity refers to the varaible phenotypes associated with an allele. There is no confusion in Mendel's mind about which allele is variably expressed since the recessive allele resulted in a uniformly white phenotype. Varible expressivity is also commonplace as any phenotype is influenced by interactions with alleles of other genes. The pleiotropic phenotype of myotonic dystrophy includes mental deficiency and muscle wastage coupled with varible expressivity since the severity of any aspects of the phenotype can be more or less severe .

Penetrance

Virtually all genotypes used by Mendel were completely expressed. All homozygous recessive ww genotypes produced the expected wrinkled seed . Similarly 100 dominant Ww genotypes resulted in phenotypically round seed. If this was not the case his F2 data would not have closely approximated the theoretical 3:1 or 9:3:3:1 ratios. When all genotypes express the expected phenotypes the phenotype is said to be 100% penetrant. The penetrance of Mendel's traits was 100%. If only 75 of 100 ww genotypes were wrinkled the penetrance would have been 75%. There are many examples of partial penetrance in humans. These include the 50-80% penetrance of neurofibromatosis and the 50-80 % penetrance of brachydactyly. The dominant inheritance of brachydactyly seen in the following pedigree readily identifies heterozygous carriers.

Lack of penetrance is seen in ? .


 

Effect of penetrance and expressivity on phenotype
genetic condition
phenotypes
Mendelian expectations
                 
variable penetrance
                 
variable expressivity
                 
variable penetrance & expressivity
                 

 


ABO Problems

Monohybrid Cross Problems: do questions 9-13


Modified Dihybrid Ratios

The classical Mendelian phenotypic ratio expected from a dihybrid AaBb is 9:3:3:1 . This phenotypic ratio resulted from assuming: (i) only two alleles of two different genes, A and B, affecting two different traits and (ii) complete dominance of one allele over the complementary recessive allele. Neither of these conditions need apply and this sets the stage for manifold modifications of the classical F2 ratio. Nevertheless, the Mendelian basis for modified ratios is secure; indeed, we can only interpret deviations by employing the classical ratio as a scaffold for the creative construction of new F2 ratios.

Every genotype can have a unique phenotype

Imagine incomplete dominance for the seed shape and seed color alleles. WW is round and ww is wrinkled but Ww is oval. GG is yellow and gg is green but Gg is orange. Selfing WwGg still follows the Laws of Segregation and Independent Assortment so that four gametes ( WG, Wg, wG, wg) produced in equal frequency combine at random to form progeny. Our newly defined dominance relationships result in identical genotypic and phenotypic ratios of 4:2:2:1:2:1:2:1:1. Find the nine components of the modified F2 ratio below- the 4/16 oval orange are filled in..

F2 gametes 1/4 WG 1/4 Wg 1/4 wG 1/4 wg
1/4 WG ? ? ? oval orange
1/4 Wg ? ? oval orange ?
1/4 wG ? oval orange ? ?
1/4 wg oval orange ? ? ?

 

A plethora of modified ratios

It is an old adage in genetics that no gene acts by itself. Humans have been estimated to have ca 80,000 genes whose coordinated and temporal action results in the final phenotype. Each gene can have many alleles which may have complex and variable dominance relationships. In addition, several different genes may have an interactive effect on the same phenotype thus masking the classical ratios. Certain 'common' patterns of modified ratios have been assigned a specific terminology - as geneticists are wont to do. Two common modifications are detailed below.

Complementary gene interaction

When William Bateson crossed two white-flowered true-breeding sweet pea lines he got a F1 with only purple flowers. The F2 segregated purple : white in a 9:7 ratio.

Recall that Mendel's cross of white by white produced only white flowers and that his cross of white by purple flowers showed a purple F1 which segregated 3 purple :1 white in the F2. How can we reconcile the different segregation patterns observed by Mendel and his 20th century disciple Bateson?

Mendel's F2 clearly results from self pollination of a monohybrid F1. If we designate the segregating recessive allele as b, then the F1 genotype is ? . Recall that although we ignore the rest of the genotype we are aware that many thousands of genes are present as homozygous or heterozygous alleles. Mendel inadvertently, or perhaps quite deliberately, selected from his 34 varieties one gene in two allelic forms that affected flower color. That is, his white line was bb and his purple line was BB. However, let's imagine that another gene "A" gene ( locus) was also involved in flower color and that Mendel's white and purple lines were homozygous dominant for this allele.Then we would designate F1 genotype as ? . Gametic and F2 expectations are given below; fill in the missing gametes.

Mendel's F2 gametes

1/2 ? 1/2 ?
1/2 ?    
1/2 Ab    
 
Bateson's F2 appears to be segregating as a dihybrid with some modification of the classical F2 ratio. 9/16 of the progeny are grouped into the purple class and 7/16 are grouped together in the white class. One way to explain Bateson's results is to assume that his white parental lines were homozygous for two different genes, "A" and "B", both involved in flower color. If we designate P1 as homozygous recessive for gene "A" the genotype for P2 is ? . Complete the table below to fully explain Bateson's F2.
Bateson's F2 gametes 1/4 ? 1/4 ? 1/4 aB 1/4 ?
1/4 ?        
1/4 Ab        
1/4 ?        
1/4 ?        

 

Possible mode of gene action:
colorless precursor-> Gene A colorless intermediate product-> Gene B -> purple pigment

or

colorless precursor-> Gene B colorless intermediate product-> Gene A -> purple pigment

 

Genetic Complementation:

We just used one of the most important tools of the geneticist! - the complementation test. The complementation test is used to decide if two mutants with the same phenotype result from a defect in the same or different genes.

1.When the hybrid of two identical phenotypes still has the same phenotype, genetic complementation is said not to have occurred and the genotypes are treated as alleles of the same? or different?

2.When the hybrid of two identical phenotypes has the wild type phenotype, genetic complementation is said to have occurred and the genotypes are treated as alleles of the same ? or different ? genes.

Epistasis

When an allele of one gene hides the effects of another gene, epistasis has occurred. The allele that is causing the masking is epistatic to the gene that it masks. Epistatic allelles can be dominant or recessive and in some cases epistasis can be reciprocal; e.g., contemplate duplicate-recessive epistasis.

A series of matings between two blood type ABs produced the following progeny: 3/16 type A; 6/16 type AB; 3/16 type B and 4/16 type O. How can we interpret this as a modified F2 ratio?

Multiple allelism of the ABO system specifies that blood type AB must be of genotype IAIB. F2 segregation for blood type would normally be expected to be 1/4 (IAIA): ? (IAIB): ? (IBIB).Our understanding of ABO multiple allelism and the Laws of Segregation and Independent Assortment makes us confident of the expected F2 segregation. The 3:6:3:4 F2 ratio must result from the epistatic interaction of the alleles of another gene.

We now must imagine the action of a second gene on the 1/4 A, 1/2 AB and 1/4 B phenotypes that would result in the observed 3:6:3:4 F2 ratio. After considerable thought we suggest another locus "H" whose recessive allele prevents the appearance of either the IA or IB allele - how else could we account for the 4/16 type O in the F2 ! ? Futhermore we suggest that the genotype of the original type AB parents is IAIBHh. Did our imagination serve us well? The table below has the answer.

F1 gametes 1/4 IAH 1/4 IAh 1/4 IBH 1/4 IBh
1/4 IAH type ? type ? type ? type ?
1/4 IAh type ? type ? type ? type ?
1/4 IBH type ? type ? type ? type ?
1/4 IBh type ? type ? type ? type ?

If the above seems confusing, see: "ABO-HH" overview : for the excellent treatment of this genetic system by Robert J Huskey.

 


Summary of Mendelian Extensions

Mendel
Extension

Heterozygous Phenotype

 

F2 variations
complete dominance

incomplete dominance

codominance

heterozygote different than either homozygote
phenotypic ratio of 1 : 2 : 1
two alleles
multiple alleles
hierarchy of phenotypes
series of 3 : 1 ratios

 

two viable alleles
lethal alleles
recessive/dominant
2 : 1 replaces 3 : 1
F2 = 9:3:3:1
gene interaction
various depending on gene action
4:2:2:1:2:1:2:1:1;   9:7;      6:3:3:4 ;      etc

Problems

1. Gene interaction Mendrill Gene interactions are recognized by apparent deviations from classical monohybrid or dihybrid Mendelian ratios. Recall the Mendelian P1 X P2 ---> F1--->F2 expectations. Starting with a cross in repulsion and using Monet's palette for the phenotype: P1 - AAbb; P2 - aaBB

P1 X P2
gametes 1/2 Ab
1/2 aB
   
F1 X F1
gametes 1/4 AB 1/4 Ab 1/4 aB 1/4 ab
1/4 AB
   
   
   
   
1/4 Ab
   
   
   
   
1/4 aB
   
   
   
   
1/4 ab
   
   
   
   

Whoops, Monet goofed - the phenotype of genotype ? is wrong. Monet was no geneticist, but he had other talents.

The F2 gene interaction table below highlights five types of gene action: 1. incomplete dominance - recessive epistasis, 2. classica Mendelianl, 3. complementary, 4. dominant epistasis and 5. recessive epistasis. Classical, complementary, and incomplete dominance - recessive epistasis interactions were discussed in our e-text. Enter these three in the correct gene interaction window below ( as 2, 3, or 1). Dominant and recessive will be presented in upcoming problems. Can you reason from the terms recessive and dominant epistasis to identify these other two modified F2 dihybrid ratios?

 

F2

genotypes

Gene interaction
? ?

?

? ?
1/16 AABB
   
  I    
2/16 AABb
2/16 AaBB
   
4/16 AaBb
1/16 AAbb
   
  II  
2/16 Aabb
1/16 aaBB
   
     
2/16 aaBb
1/16 aabb
   
II  

Modelling for complementary gene interaction, the genotype of 9/16 purple is ? . A cross of the 2/16 F2 class AABb to aabb would give ? white to ? purple.A cross of the 4/16 F2 class to aabb would give ? white to ? purple.

Modelling for the ABO alleles [A = I A , a = I B ] and the Bombay alleles [B = H, b = h], the gene interaction blood type I is type  ? and blood type II is ? .

 

 

 

The genotypic complexity underlying phenotypes is a challenging and intriguing fascination to geneticists. The classical Mendelian patterns - dependent on bi-allelles, complete dominance, 100% penetrance, no lethality, uniform expressivity and absence of gene interaction - liken a mirage surface rippled by secondary waves. The Mendelian surface is secure. Our tools for wave construction are disciplined imagination .... and perseverance.

 

2. When a pea combed chicken was crossed to a rose combed chicken the progeny were all walnut combed. The F2 segregated 9/16 walnut [R_P_], 3/16 pea, 3/16 rose and 1/16 single.The F2 genotype of: rose-: ? ; pea- ? ; single- ? . F1 walnut when crossed to F2 single would result in walnut : rose : pea : single in a ? ratio.

How is comb inheritance a type of complementary gene action similar to the 9:7 F2 dihybrid flower color ratio ?

3.Vincent van Gogh "Irises" http://metalab.unc.edu/wm/paint/auth/gogh/irises/gogh.irises.jpg

The red pigment cyanidin is synthesized from a colorless precursor. Addition of a hydroxyl group (-OH) to cyanidin results in a purple pigment. Two purple plants were crossed to give : 94 purple, 31 red, 43 white. Assign genotpyes to the purple plants and explain the progeny flower colors.

4. A cross between white leghorn http://www.ansi.okstate.edu/poultry/chickens/leghorn/whtsngl.htm and white wyandotte http://www.ansi.okstate.edu/poultry/chickens/wyandott/white.htm chickens produces all white F1 progeny. The F2 segregates 13 white : 3 colored. Explain this modified dihybrid ratio .

http://www.miceandrats.com/aov_ago.htm 5. The AY allele of the mouse agouti gene is dominant for yellow coat color but acts as a recessive lethal. cc at the albino locus produced albinism. What results from a crossing two mice heterozygous at both these loci ?

6. Four babies become accidently mixed up in a maternity ward. The ABO types of babies are : A, B, O, AB. The ABO types for the four sets of parents are :  AB X O;  A X O;   O X O;  A X AB. Match the baby with the biological parents.

http://www.thebabynet.com/contest/media http://www.thebabynet.com/contest/media http://www.thebabynet.com/contest/media

baby parents
A  ?
B  ?
O  ?
AB  ?

7. Coat color in the guinea pig may be controlled by at least four alleles. C (black), ck (sepia), cd (cream), ca (albino) where C > ck > cd > ca.

cream

http://www.users.wineasy.se/dan.johansson/eva/mars/ras/self.htm

black

(i) For the cross P1 sepia x P2 cream, when sepia and cream both had an albino parent, determine the phenotypic ratios. ?

(ii) For the cross P1 sepia x P2 cream, when sepia had an albino parent and cream had two sepia parents, determine the phenotypic ratios.

Sepia has three possible genotypes : ckck, ckcd, ckca. Cream has two possible genotypes : cdcd, cdca. P1 parent is : ckca since one allele must be ck (from ck_ parent) and one allele must be ca (from caca parent). P2 parents can be ckcd or ckca, so P2 can be cdcd or cdca. ?

8. This deaf-mutism problem is modified from one of Dr. Robert Huskey's pages.

http://wsrv.clas.virginia.edu/~rjh9u/deafmute.html

The pedigree above demonstrates four generations of deaf-mutism (DM) in families of northern Ireland. The family to the right of the dashed line shows segregation for an autosomal recessive form of DM as clearly illustrated by the mating of I-3 and I-4 who produce all affected children.

The family to the left of the dashed line is also segregating for an autosomal recessive form of DM although the parents II-1 and II-2 must both be heterozygous for the trait.

The curiousity is that although III-8 (from the left family) and III-9 (from the right family) each have the DM trait, none of their children show the trait of deaf-mutism!

Given the genotypes of I-1 and I-3 ? , fill in the remaining genotypes.

9. Blue Sclera http://www.nejm.org/content/1998/0339/0014/0966.asp Bone Fractures http://medgen.genetics.utah.edu/photographs/pages/osteogenesis_imperfecta.htm

Osteogenesis imperfecta is caused by a dominant allele COL1A1 and is 100% penetrant for brittle bones. However other aspects of the pleiotropic phenotype such as hearing loss and blue sclera have variable expressivity. The following was modified from OMIM and indicates the type of complex variable phenotypes associated with genetic diseases - and other phenotypes!

Osteogenesis imperfecta is characterized chiefly by multiple bone fractures, usually resulting from minimal trauma. Affected individuals have blue sclerae, normal teeth, and normal or near-normal stature. Hearing loss of conductive or mixed type occurs in about 50% of families, beginning in the late teens and leading, gradually, to profound deafness, tinnitus, and vertigo by the end of the fourth to fifth decade.

The disorder may exhibit considerable interfamilial and intrafamilial variability in the number of fractures and degree of disability. Most striking were identical twins, the offspring of a mildly affected mother. Twin B was born small for gestational age, had 12 fractures and was 150 cm tall (third centile) at 11 years of age. Her twin was born appropriate for gestational age and had had only 2 fractures at age 8 and 9 secondary to strenuous exercise; her current height was 162 cm (fiftieth centile).

Willing et al., 1990 noticed that the more severely affected family members had children with both mild and severe phenotypes, while the mildly affected individual had an offspring with a mild phenotype. This suggested to them that there might be some other, not identified, factor segregating independently in this family that acts to modulate the final phenotype.

The pedigree follows the inheritance of COL1A1 and the resultant phenotypes for brittle bones, blue sclera and deafness.

Assuming 100% penetrance for brittle bones the genotype for all 14 filled in individuals is ? . Blue sclera shows ? % penetrance and deafness is ? % penetrant.

 

10. Interacting biochemical pathways provide an apt model of the genetic mechanic. Many of the tens of thousands of genes in a eukaryotic species participate in biochemical pathways whose products merge and interact to craft the phenotype. In the following problem three biochemical pathways interact. A, B, C, D, and E are completely dominant alleles resutling in the corresponding biochemicals A - yellow, B - red, C - yellow, D - white, and E -blue. Recessive alleles a, b, c, d, and e result in the absence of colored compounds. Imagine colors represent spots on a pelicans pouch, or what ever suits your fancy. Pathway III normally does not influence color but if D is recessive, white3 builds up in concentration and can be converted to yellow by pathway IV. Caveats are: red + blue = purple ; blue + yellow = green ; white is colorless - that is, white is white.

Pathway I -----> white1 ---(E)------> blue

Pathway II ----> white2 ---(A)------>yellow---(B)--> red

Pathway III - - - -------------------> white3 ---(D)---> white 4

Pathway IV ------white3 of pathway III----(C)---->yellow of pathway II.

Explain the following F2s derived from pure breeding parentals.

(i) 9 purple : 3 green : 4 blue ?

(ii) 9 purple : 3 red : 3 blue : 1 white ? --> Lab 4 exercise

(iii) 13 purple : 3 blue ? --> Lab 4 exercise

(iv) 9 purple : 3 red : 3 green : 1 yellow ? -> Lab 4 exercise

A difficult problem. But imagine how complexity spirals with gene numbers >104, alleles>102, and rampant interactions that oscillate with time and the environment!   How Mendel ! ...... Hearken Stern and Sherwood.

 

 

11.

http://der.org/docued/films/moon-children.html Albinism is inherited as the recessive alleles of three different genes OCA1, OCA2 and OCA3. The genes are fully complementary so that crosses between OCA1OCA1 X OCA2OCA2 is pigmented wild type ? OCA1+, OCA2 ? . In addition to lack of pigmentation, the pleiotropic phenotype can include impaired vision. Assume that homozygosity at any two recessive OCA genes will reduce vision by 25% and that homozygotes for all three OCA genes will be blind. If the paternal genotype is OCA1OCA1, OCA2OCA2+, OCA3OCA3+ and the maternal genotype is OCA1OCA1+, OCA2OCA2+, OCA3OCA3.

(i) what fraction of albino children will have vision reduced 25% ? Concept problem 2

(ii)What fraction of childen will be albino and blind ? Concept problem 2  

12. How might Mendel have responsed to earlier questions posed by William Blake (1759-1827) in his poem "The Tyger". http://www.ngv.vic.gov.au/blake/images/tyger1.jpg Jenny Fowler's Answer


Additional Problems: do problems 9-13


Rembrandt " The Anatomy Lesson of Dr. Nicolaes Tulp"

Social and Ethical Issues http://metalab.unc.edu/wm/paint/auth/rembrandt/1630/nicolaes-tulp.jpg

Hemochromatosis is a late-onset recessive genetic disease estimated to affect 1 in 200 North Americans. The cost to society for medical treatment of individuals with advanced hemochromatosis is a major concern. A recent column by Ann Landers reports that a genetic test can identify homozygous HH children in time to employ preventive therapy. Phlebotomy (bloodletting) is a simple inexpensive therapy that could be used to prevent millions of childrens' future suffering. Full implementation of this therapy requires the early use of the HH allele diagnostic test prior to onset of the clinical symptoms of hemochromatosis. How should society react to this genetic technology? Should all children be screened and parents advised of the results? If so, do parents have the right to refuse treatment of their children? How does your response compare to public-interest legislation such as wearing a seat belt or prevention of smoking in public places?


Waterhouse "Destiny"

What Lies Ahead? http://metalab.unc.edu/wm/paint/auth/waterhouse/destiny.jpg

Cystic fibrosis is said to be one of the most common single gene diseases in the world, affecting 1 in 2000 children. CF alleles act as late-onset lethals since most patients die from severe lung infections by 30 years of age. Although more than 800 recessive CF alleles have been identified, likely not all variations in the development of the disease are due to different allelic phenotypes. Given the wide range of known gene interations in other genetic systems it is believed that many clinical variations resulted from the modifying effect of other genes on the CFCF genotype.

In June of 1999, Dr. Julian Zielenski of The Hospital for Sick Children and the University of Toronto announced the identification of a gene that modifies the severity of cystic fibrosis.This gene, which acts on an intestinal obstruction, can effect up to 20% of cystic fibrosis patients. The identification of the modifing gene in humans was made possible by earlier research with a similar CF modifying gene in mice.

The press release opined that " The identification of modifier genes in CF will allow scientists to gain a better understanding of the different clinical presentations of the disease which, it is anticipated, will lead to insights into prognosis and management of cystic fibrosis, as well as development of novel therapies. "

Given the eminent identificaion of all ca 80,000 human genes, we can look forward to a day when, in addition to identification of major disease causing genes, genes that affect the severity of disease genes will also be identified.


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