台大農藝系遺傳學 601 20000 chapter 12 slide 1 chapter 13 extensions of mendelian genetic...

台大農藝系遺傳學 601 20000 Chapter 12 slide 1

CHAPTER 13Extensions Of Mendelian

Genetic Analysis

Peter J. Russell

edited by Yue-Wen Wang Ph. D.Dept. of Agronomy, NTU

A molecular Approach 2nd Edition


Determining the Number of Genes for Mutations with the Same Phenotype

1. Relationship between phenotype and gene can be studied through mutants identified by phenotype distinct from wild type.

2. Complementation test (cis-trans test) determines whether independently isolated mutations for the same phenotype are in the same or different genes by crossing two mutants (Figure 13.1).

a. If mutations are in different genes, phenotype will be wild type (complementation).

b. If mutations are in the same gene, phenotype will be mutant (no complementation).


Fig. 12.1 Complementation test to determine whether two mutations resulting in the same phenotype are in the same or different genes


3. Drosophila provides an example. Wild-type body color is grey-yellow. If two true-breeding mutant black-bodied strains are crossed, all F1 are wild type (Figure 13.2).

a. Genes are e (ebony) and b (black). Black parents are homozygous mutant but in different genes (e/eb+/b+) and (e+/e+b/b).

b. F1 are heterozygous at both loci (e+/eb+/b) and therefore wild type, showing complementation.


Multiple Alleles

1. Not all genes have only two forms (alleles); many have multiple alleles (figure 13.3). No matter how many alleles for the gene exist in the multiple allelic series, however, a diploid individual will have only two alleles, one on each homologous chromosome.

台大農藝系遺傳學 601 20000 Chapter 12 slide 6Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings.

Fig. 13.3 Allelic forms of a gene


ABO Blood Groups1. ABO blood groups are important in blood transfusions, and result from a series

of three alleles (IA, IB and i) that combine to produce four phenotypes (A, B, AB and O).

2. Both IA and IB are dominant to i, while IA and IB are codominant to each other. The resulting phenotypes are (Table 13.1):

a. People with genotype i/i are blood type O.

b. People with genotype IA/IA or IA/i are blood type A.

c. People with genotype IB/IB or IB/i are blood type B.

d. People with genotype IA/IB are blood type AB.

3. ABO inheritance follows Mendelian principles. For example, a type O individual’s genotype is i/i. Possible genotypes of the parents could be:

a. i/i and i/i (both blood type O).

b. IA/i and i/i (one type A, the other type O).

c. IA/i and IA/i (both type A).

d. IB/i and i/i (one type B, the other type O).

e. IB/i and IB/i (both type B).

f. IA/i and IB/i (one type A, the other type B).

4. Blood typing may be used in cases of disputed parentage. Blood typing does not prove the identity of a parent. It can, however, eliminate individuals who are not biological parents of a particular child. Example:

a. A child with blood type AB (IA/IB) could not have a parent with type O (i/i).

b. Blood type data are not considered adequate legal proof for parenthood in most states, and DNA fingerprinting is generally used.

iActivity: Was She Charlie Chaplin's Child?


Biochemical Genetics of the Human ABO Blood Group1. Cellular antigens are important in blood transfusions, since recipient antibodies

may respond to antigens on donor cells.

2. Karl Landsteiner discovered human ABO blood groups in the early 1900s, and received the 1930 Nobel Prize in Physiology or Medicine for this work. Properties of the human ABO blood group:

a. There are three alleles at the ABO locus, IA, IB, and i. From these three alleles, four phenotypes are produced:

i. Type A individuals have the A antigen on their RBCs, and anti-B antibodies in their blood. Their genotype is IA/IA or IA/i.

ii. Type B individuals have the B antigen on their RBCs, and anti-A antibodies in their blood. Their genotype is IB/IB or IB/i.

iii. Type AB individuals have both the A and the B antigen on their RBCs, and neither anti-A nor anti-B antibodies in their blood. Their genotype is IA/IB.

iv. Type O individuals have neither the A nor the B antigen on their RBCs, and both anti-A and anti-B antibodies in their blood. Their genotype is i/i.

3. Antigen–antibody relationships, and their impact in blood transfusions, are summarized in Figure 13.4.


Fig. 13.4 Antigenic reactions that characterize the human ABO blood types


4. Summary of the relationship between the ABO alleles and RBC antigens:

a. The ABO locus produces RBC antigens by encoding glycosyltransferases, which add sugars to existing polysaccharides on membrane glycolipid molecules. These polysaccharides act as the antigen in the ABO system (Figure 13.5).

b. In most people, the glycolipid is the H antigen.

i. Activity of the IA gene product, a-N-acetylgalactosamyl transferase, converts the H antigen to the A antigen.

ii. Activity of the IB gene product, a-D-galactosyltransferase, converts the H antigen to the B antigen.

iii. Both enzymes are present in an IA/IB individual, and some H antigens will be modified to the A antigen while others are modified to the B antigen.

iv. Neither enzyme is present in an i/i individual, and so the H antigen remains unmodified.

5. Production of the H antigen is controlled by a different genetic locus from the ABO enzymes. Rarely, an individual lacks the dominant allele H needed for H antigen production. This h/h genotype results in the Bombay blood type, which is similar to type O except that Bombay blood type individuals produce anti-O antibodies that are not seen in true type O individuals.


Drosophila Eye Color1. Drosophila has over 100 mutant alleles at the eye-color locus on

the X chromosome. Example designations for alleles at this locus:a. The white-eyed variant allele is designated w.

b. The wild-type (brick red) allele is w+ .

c. A recessive allele, we, produces eosin (reddish-orange) eyes.2. Soon after Morgan’s discovery of X-linkage, he found new genes for eye shape.

and color, including a red one called vermilion (v+).

a. He experimentally crossed a white-eyed female with a vermilion eyed male.

b. The F1 females were all red-eyed (wild type), rather than either vermilion or white.

c. He concluded two different genes were involved in Drosophila eye color (white and vermilion) rather than just alleles of a single locus.

d. The original cross was: w v+/ w v+ (white-eyed female) with w+ v/ Y (vermilion-eyed male).

e. The F1 females (wild-type red eyes) would be w v+/ w+ v, doubly heterozygous.

3. The eosin allele is recessive to wild-type. Morgan (1912) crossed an eosin-eyed female with a white-eyed male. All F1 females had eosin eyes.


4. Sturtevant (1913) concluded that eosin and white are mutations of a single gene. The relationship between these multiple alleles is:

a. The allele red (wild-type) is dominant to eosin and white.

b. The eosin allele is recessive to red, but dominant to white.

c. For example, in the cross of an eosin-eyed (we/ we) female. with a white-eyed male (w/Y), the Fl females are all we/w. They have eosin eyes, showing that we is dominant over w.

d. Next the eosin-eyed Fl females (we/w) are crossed with red-eyed males w+/ Y (Figure 13.6).

(1) All female progeny are red-eyed (w+/w or w+/we).

(2) Male progeny are 1/2 eosin-eyed (we/Y), and 1/2 white-eyed (w/Y).

e. Many alleles of the white-eye gene exist, producing a wide range of colors depending on the deposition of pigment in the eye cells (Table 13.2).


Fig. 13.6 Results of crosses of Drosophila melanogaster involving two mutant alleles

of the same locus


5. The number of possible genotypes in a multiple allelic series depends on how many alleles are involved (Table 13.3).

a. The formula n(n + 1)/2 calculates possible genotypes for n alleles.

b. Of the genotypes predicted by the formula, n are homozygotes, and n(n - 1)/2 are heterozygotes.


Relating Multiple Alleles to Molecular Genetics

1. Genes encode proteins, and changes in amino acids of those proteins may change a phenotype. Multiple alleles exist for many genes, because there are many sites within a gene where introduction of a mutation will alter the protein product.

2. Consequences of multiple alleles in human genetic disorders include:

a. Variation in disease symptoms depending on the patient’s allele(s).

b. Complications in designing a single DNA-based test to diagnose the disease or detect carriers.


Modifications of Dominance Relationships

1. Complete dominance and complete recessiveness are two extremes in the range of dominance possible between pairs of alleles. Many allelic pairs are less extreme in their expression, showing incomplete dominance or codominance.a. In incomplete dominance, a heterozygote’s phenotype

will be intermediate between the two possible homozygous phenotypes.

b. In codominance, the heterozygote shows the phenotypes of both homozygotes.

c. At the molecular level, these relationships between pairs of alleles depend upon patterns of gene expression.


Incomplete Dominance

Animation: Incomplete Dominance and Codominance

1. Incomplete dominance is an allelic relationship where dominance is only partial. In a heterozygote, the recessive allele is not expressed. The one dominant allele is unable to produce the full phenotype seen in a homozygous dominant individual. The result is a new, intermediate phenotype.

2. An example is plumage color in chickens (Figure 13.7):a. Crossing a true-breeding black chicken (CBCB) with a true-

breeding white one (CWCW) produces an Andalusian blue F1 (CBCW).

b. When the F1 interbreed, the F2 include black (CBCB), Andalusian blue (CBCW) and white (CWCW) birds, in a ratio of 1:2:1.

c. At the molecular level, two copies of CB produce black, while 1 copy is sufficient to produce only the gray “Andalusian blue” phenotype.


Fig. 13.7 Incomplete dominance in chickens


3. Palomino horses (golden-yellow body with nearly white mane and tail) are another example (Figure 12.4). When palominos are interbred, the progeny are:

a. 1/4 cremello (cream colored) with genotype Ccr/Ccr.

b. 1/2 palomino with genotype C/ Ccr.

c. 1/4 light chestnut with genotype C/C.

4. Incomplete dominance often occurs in plants. An example is flower celor in snapdragons involving two alleles, CR and CW. Red-flowered plants (CR/CR) crossed with white-flowered ones (Cw/Cw) produce all pink progeny (CR/Cw ).


Codominance

1. In codominance, the heterozygote’s phenotype includes the phenotypes of both homozygotes. Examples include:a. The ABO blood series, in which a heterozygous IA/IB

individual will express both antigens, resulting in blood type AB.

b. The human M-N blood group involves red blood cell antigens that are less important in transfusions. There are three types:

i. Type M, with genotype LM/LM.

ii. Type MN, with genotype LM/LN.

iii. Type N, with genotype LN/LN.


Molecular Explanations of Incomplete Dominance and Codominance

1. Current explanations involve levels of gene expression for each allele in the pair.a. In codominance, both alleles make a product, producing a

combined phenotype.

b. In incomplete dominance, the recessive allele is not expressed, and the dominant allele produces only enough product for an intermediate phenotype.

c. By contrast, a completely dominant allele creates the full phenotype by one of two methods:

i. It produces half the amount of protein found in a homozygous dominant individual, but that is sufficient to produce the full phenotype. These genes are haplosufficient.

ii. Expression of the one active allele may be upregulated, generating protein levels adequate to produce the full phenotype.


Gene Interactions and Modified Mendelian Ratios1. Phenotypes result from complex interactions of molecules

under genetic control. Genetic analysis can often detect the patterns of these reactions. For example:a. In the dihybrid cross A/a B/b X A/a B/b, nine genotypes will result.

b. If each allelic pair controls a distinct trait and exhibits complete dominance, a 9:3:3:1 phenotypic ratio results.

c. Deviation from this ratio indicates that interaction of two or more genes is involved in producing the phenotype.

2. Two types of interactions occur:a. Different genes control the same general trait, collectively

producing a phenotype.

b. One gene masks the expression of others (epistasis) and alters the phenotype.

3. Examples here are dihybrid, but in the “real world” larger numbers of genes are often involved in forming traits.

4. The molecular explanations offered here are currently hypothetical models, and await rigorous analysis using the tools of molecular biology.


Gene Interactions That Produce New Phenotypes

1. Nonallelic genes that affect the same characteristic may interact to give novel phenotypes, and often modified phenotypic ratios. Examples include:a. Comb shape in chickens, influenced by two gene loci to produce four

different comb types. Each will breed true if parents are homozygous (Figure 13.8).

i. In a cross between a homozygous rose-combed (R/R p/p) bird and a single-combed (r/r p/p) bird (Figure 13.9):

(1) The F1 (R/r p/p) will all have rose combs.

(2) The F2 will be 3 rose (R/– p/p) : 1 single (r/r p/p).

ii. Similarly, pea comb (r/r P/P) is dominant over single (r/r p/p), with F1 (r/r P/p) all showing pea combs, and a 3:1 ratio of pea to single in the F2.

iii. Crossing true-breeding rose (R/R p/p) and pea (r/r P/P) results in:

(1) An F1 with all walnut combs (R/r P/p).

(2) An F2 showing a ratio of 9 walnut (R/– P/–) : 3 rose (R/– p/p) : 3 pea (r/r P/–) : 1 single (r/r p/p).

iv. These interactions fit the expected ratios for a Mendelian dihybrid cross. The molecular basis for each phenotype is unknown, but it appears that the dominant alleles R and P each produce a factor that modifies comb shape from single to a more complex form.


Fig. 13.8


Fig. 13.9 Complete dominance in chickens


b. Fruit shape in summer squash shows a 9:6:1 ratio. Two genes are involved, each completely dominant. Interaction between the two loci produces a new phenotype (Figure 13.10).

i. Long fruit (a/a b/b) are always true-breeding.

ii. Sphere-shaped fruit (A/– b/b or a/a B/–) are not always true-breeding, and sometimes produce long (a/a b/b) or disk-shaped (A/– B/–) fruit.

iii. A cross between true-breeding spherical strains (A/A b/b x a/a B/B) produces a disk-shaped F1. The F2 will be 9⁄16 disk-shaped (A/– B/–), 6⁄16 spherical (A/– b/b or a/a B/–) and 1⁄16 long (a/a b/b). This modification of the Mendelian ratio indicates that two loci are involved.

iv. The precise molecular basis of these phenotypes is unknown.


Fig. 13.10 Generation of an F2 9:6:1 ratio for fruit shape in summer squash


Epistasis

1. In epistasis, one gene masks the expression of another, but no new phenotype is produced.a. A gene that masks another is epistatic.

b. A gene that gets masked is hypostatic.

2. Several possibilities for interaction exist, all producing modifications in the 9:3:3:1 dihybrid ratio:a. Epistasis may be caused by recessive alleles, so that a/a

masks the effect of B (recessive epistasis).

b. Epistasis may be caused by a dominant allele, so that A masks the effect of B.

c. Epistasis may occur in both directions between genes, requiring both A and B to produce a particular phenotype (duplicate recessive epistasis).


3. In recessive epistasis, A/-b/b and a/ab/b have the same phenotype, producing an F2 ratio of 9;3;4. Examples include:

a. Coat color determination in rodents (Figures 13.11 and 13.12)

i. Wild mice have individual hairs with an agouti pattern, bands of black (or brown) and yellow pigment. Agouti hairs are produced by a dominant allele, A. Mice with genotype a/a do not produce yellow bands and have solid-colored hairs.

ii. The B allele produces black pigment, while b/b mice produce brown pigment. The A allele is epistatic over B and b, in that it will insert bands of yellow color between either black or brown.

iii. The C allele is responsible for development of any color at all, and so it is epistatic over both the agouti (A) and the pigment (B) gene loci. A mouse with genotype c/c will be albino, regardless of its genotype at the A and B loci.

iv. In the cross A/aC/c´A/aC/c, the offspring will be:

(1) 9⁄16 agouti (A/-C/-).

(2) 3⁄16 solid (a/aC/-).

(3) 4⁄16 albino (3⁄16 A/-c/c+1⁄16 a/ac/c).


Fig. 13.11 Recessive epistasis: Generation of an F2 9:3:4 ratio for coat color in rodents


b. Coat color determination in labrador retriever dogs (Figure 13.12)

i. Gene B/- makes black pigment, while b/b makes brown.

ii. Another gene, E/- allows expression of the B gene, while e/e does not.

iii. Genotypes and their corresponding phenotypes:

(1) B/-E/- is black

(2) b/bE/- is brown (chocolate)

(3) -/-e/e produces yellow with nose and lips either dark (B/-e/e) or pale (b/be/e)


4. In dominant epistasis A/-B/- and A/-b/b have the same phenotype, producing an F2 ratio of 12;3;1. Examples:

a. Summer squash fruit have 3 common colors, white, yellow, and green (Figure 13.13).

i. Yellow is recessive to white but dominant to green.

ii. Gene pairs are W/w and Y/y.

(1) W/- are white no matter the genotype of the other locus.

(2) w/w are yellow in Y/- and green in y/y.

iii. Theoretical biochemical pathway is shown in Figure 13.14, postulating that a white molecule is converted to a green intermediate and then to a yellow end product.

(1) Dominant Y required to convert green intermediate to yellow.

(2) Dominant W inhibits white-to-green conversion, resulting in white fruit.

b. Greying in horses is an example of epistasis that occurs slowly (Figure 13.15).

i. The dominant allele G results in progressive silvering of the coat from birth to maturity, masking expression of other coat color genes.

ii. Horses with g/g retain the coat color they had at birth.


Fig. 13.13 Dominance epistasis


Fig. 13.14 Dominance epistasis


5. Duplicate recessive epistasis (complementary gene action) involves two loci that each produce an identical phenotype, showing a 9;7 ratio. Example is flower color determination in sweet peas (Figure 13.16).

a. Purple is dominant for flower color, and when a true-breeding purple plant is crossed with a true-breeding white one, the F2 shows a typical 3;1 ratio.

b. White strains usually breed true, but occasionally the cross of two different white strains (p/pC/C´P/Pc/c) will produce an F1 that is entirely purple (P/pC/c).

i. The F2 of this cross will be 9⁄16 purple (P/-C/-), and 7⁄16 white (3⁄16 P/-c/c+3⁄16 p/pC/-+1⁄16 p/pc/c).

ii. All of the white F2 plants will breed true, as will 1⁄9 of the purple F2 plants (P/PC/C).

c. Two genes appear to be involved. The C/c alleles determine whether the flower can have color, and the P/p alleles determine whether purple is produced.


Fig. 13.16 Duplicate recessive epistasis: Generation of an F2 9:7 ratio for flower color

in sweet peas


6. Duplicate dominant epistasis shows a 15;1 ratio, and is illustrated by fruit shape in Shepherd’s purse plant.

a. There are two possible fruit shapes, heart-shaped and narrow.

b. One dominant allele of either duplicate gene will produce heart-shaped fruit.

i. A/-B/-, A/-b/b, and a/aB/- all produce heart-shaped fruit.

ii. a/ab/b produces narrow fruit.

7. Interactions between genes can produce many types of phenotypes. They are detected by deviations from expected phenotypic ratios. Table 13.4 shows examples.

8. The complex relationships of epistasis play a role in many human genetic disorders, further complicating their analysis.


Essential Genes and Lethal Alleles1. Some genes are required for life (essential genes), and

mutations in them (lethal alleles) may result in death. Dominant lethal alleles result in death of both homozygotes and heterozygotes, while recessive lethal alleles cause death only when homozygous.

2. An example is the yellow body color gene in mice (Cuenot, 1905):

a. Yellow crossed with nonyellow results in a ratio of 1 yellow; 1 nonyellow. This suggests yellow is heterozygous.

b. Yellow mice never breed true, another indication of heterozygosity. When yellow is bred with yellow, the result is about 2 yellow; 1 nonyellow (instead of the predicted 3;1).

c. Castle and Little (1910) proposed that yellow homozygotes die in utero, and are therefore missing from the progeny. The yellow allele has a dominant effect on coat color, but also acts as a recessive lethal allele.

d. Yellow is an allele of the agouti locus, designated AY. Figure 13.17 shows the yellow 3 yellow cross.

i. The cross is AY/A+´AY/A+, and death of the homozygous yellow animals (AY/AY) results in a 2;1 ratio.

ii. When two heterozygotes are crossed and produce a 2;1 ratio of progeny, a recessive lethal allele is suspected.


Fig. 13.17 Inheritance of a lethal gene in mice


e. Molecular cloning of the agouti locus assists in analysis of these phenotypes:

i. Wild-type agouti mice express the agouti gene only during hair development in the days after birth, and when plucked hair is being regenerated. Gene expression is seen in no other tissues and at no other time.

ii. Heterozygous mice (AY/A+) express the AY allele at high levels in all tissues during all developmental stages. Tissue-specific regulation appears to be lost in the AY allele.

iii. The AY allele transcript RNA is 50% longer than that of the wild-type allele (A+). This is because:

(1) The AY allele results from deletion of an upstream sequence, removing the normal promoter of the agouti gene.

(2) The gene is transcribed from the promoter of an upstream gene called Raly. The beginning of the sequence encoding Raly is fused with the agouti gene, producing a longer transcript.

iv. Embryonic lethality of AY/AY mice probably results from lack of Raly gene activity, rather than from the defective agouti gene.


3. Essential genes probably occur in all diploid organisms, including snapdragons and Drosophila.

4. Human examples of recessive lethal alleles include:

a. Tay-Sachs disease, resulting from an inactive gene for the enzyme hexosaminidase. Homozygous individuals develop neurological symptoms before 1 year of age, and usually die within the first 3–4 years of life.

b. Hemophilia results from an X-linked recessive allele, and is lethal if untreated.

5. A dominant lethal gene causes Huntington disease, characterized by progressing central nervous system degeneration. The phenotype is not expressed until individuals are in their 30s. Dominant lethals are rare, since death before reproduction would eliminate the gene from the pool.


Gene Expression and the Environment

1.Development of a multicellular organism from a zygote is a series of generally irreversible phenotypic changes resulting from interaction of the genome and the environment. Four major processes are involved:a.Replication of genetic material.

b.Growth.

c.Differentiation of cells into types.

d.Arrangement of cell types into defined tissues and organs.

2.Internal and external environments interact with the genes by controlling their expression and interacting with their products.


Penetrance and Expressivity1. Penetrance describes how completely the presence of an

allele corresponds with the presence of a trait. It depends on both the genotype (e.g., epistatic genes) and the environment of the individual (Figure 13.19).

a. If all those carrying a dominant mutant allele develop the mutant phenotype, the allele is completely (100%) penetrant.

b. If some individuals with the allele do not show the phenotype, penetrance is incomplete. If 80% of individuals with the gene show the trait, the gene has 80% penetrance.

c. Human examples include:

i. Brachydactyly involves abnormalities of the fingers, and shows 50–80% penetrance.

ii. Many cancer genes are thought to have low penetrance, making them harder to identify and characterize.


2. Expressivity describes variation in expression of a gene or genotype in individuals.

a. Two individuals with the same mutation may develop different phenotypes, due to variable expressivity of that allele.

b. Like penetrance, expressivity depends on both genotype and environment, and may be constant or variable.

c. A human example is osteogenesis imperfecta, inherited as an autosomal dominant with nearly 100% penetrance.

i. Three traits are associated with the allele:

(1) Blueness of the sclerae (whites of eyes).

(2) Very fragile bones.

(3) Deafness.

ii. Osteogenesis imperfecta shows variable expressivity, because an individual with the allele may have one, two, or all three of its symptoms, in any combination. Bone fragility is also highly variable.


Fig. 13.18 Illustrations of the concepts of penetrance and expressivity in the

phenotypic expression of a genotype


3. Some genes have both incomplete penetrance and variable expressivity. An example is neurofibromatosis (Figure 13.19).a. The allele is an autosomal dominant that shows 50–80%

penetrance and variable expressivity.

b. Individuals with the allele show a wide range of phenotypes:

i. The mildest form of the disease is a few pigmented areas on the skin (café-au-lait spots).

ii. More severe cases may include:

(1) Neurofibroma tumors of various sizes.

(2) High blood pressure.

(3) Speech impediments.

(4) Headaches.

(5) Large head.

(6) Short stature.

(7) Tumors of eye, brain or spinal cord.

(8) Curvature of the spine.

4. Incomplete penetrance and variable expressivity complicate medical genetics and genetic counseling.


Effects of the Environment1. Age of onset is an effect of the individual’s internal

environment. Different genes are expressed at different times during the life cycle, and programmed activation and inactivation of genes influences many traits. Human examples include:a. Pattern baldness, appearing in males aged 20–30 years.

b. Duchenne muscular dystrophy, appearing in children aged 2–5 years.

2. Sex of the individual affects the expression of some autosomal genes.a. Sex-limited traits appear in one sex but not the other.

Examples include:

i. Milk production in dairy cattle, where both sexes have milk genes, but only females express them.

ii. Horn formation in some sheep species, where only males express the genes used to produce horns.

iii. Facial hair distribution in humans.


b. Sex-influenced traits appear in both sexes, but the sexes show either a difference in frequency of occurrence or an altered relationship between genotype and phenotype. Human examples include:

i. Pattern baldness, controlled by an autosomal gene that is dominant in males and recessive in females.

(1) The genotype b/b produces pattern baldness in both men and women.

(2) The genotype b+/b+ gives a nonbald phenotype in both sexes.

(3) The genotype b+/b will lead to the bald phenotype in men, and the nonbald phenotype in women.

ii. Cleft lip and palate (2:1 ratio of males to females).

iii. Clubfoot (2:1).

iv. Gout (8:1).

v. Rheumatoid arthritis (1:3).

vi. Osteoporosis (1:3).

vii. Systematic lupus erythematosus (1:9).


Fig. 13.20 Sex-influenced inheritance of pattern baldness in humans


3. Temperature may alter the activity of enzymes so that they function normally at one temperature but are nonfunctional at another. An example is fur color in Himalayan rabbits (Figure 13.21).a. These white rabbits develop darker fur on

the cooler parts of their bodies (ears, nose and paws).

b. Since all body cells have the same genotype, this fur pattern might result from environmental influences. This was tested by raising Himalayan rabbits under different temperature conditions:

i. Rabbits reared above 30°C were entirely white.

ii. Rabbits raised at 25°C had the typical Himalayan phenotype.

iii. Rabbits raised at 25°C with part of the body experimentally cooled to below 25°C, had a dark spot on the experimentally cooled part.


4. Chemicals can have significant effects. Two examples:a. Phenylketonuria (PKU) is an autosomal recessive defect in

metabolism of the amino acid phenylalanine. If not treated by restricting phenylalanine in the diet, severe mental retardation and other symptoms result.

b. A phenocopy is a modification of the phenotype caused by environmental conditions, mimicking a known gene mutation. Phenocopies are not hereditary, and the individual does not carry the allele(s) being mimicked. Examples of phenocopies include:

i. Rubella, which produces cataracts, deafness and heart defects in a fetus whose mother is infected during the first 12 weeks of pregnancy, mimics rare recessive alleles.

ii. The drug thalidomide, taken on days 35–50 of gestation, mimics the effects of the genetic disorder phocomelia, suppressing development of long bones in the limbs.


Nature versus Nurture1. Phenotypes seen for many traits are influenced by both genes and environment.

Some human examples:a. Human height has both genetic and environmental components.

i. Genetically, children tend to have about the same stature as their parents, and several genetic forms of dwarfism are known (achondroplasia is an example).

ii. Environmentally, diet and health care are probably responsible for the increase in human height of about 1 inch per generation over the last century.

b. Alcoholism is an example of a behavioral trait influenced by both genes and environment.

i. A genetic influence is shown in studies of adopted children. Those with alcoholic biological fathers are significantly more likely to become alcoholics than those with non-alcoholic biological fathers.

ii. Environment plays a key role also, since alcoholism can only develop if alcohol is available.

iii. Genes make individuals more or less susceptible to alcohol abuse, perhaps by affecting metabolism of alcohol or development of personality traits involved in drinking, but the genes alone do not produce the phenotype.

c. Human intelligence is an example of a very complex relationship between genes and environment.

i. Genetic disorders are known to produce mental retardation. Examples are PKU and Down syndrome. Genes also influence IQ among non-retarded people, with adopted children scoring closer to their biological parents than to their adoptive parents.

ii. Environmental influences are seen in studies of identical twins, who frequently differ in IQ scores.

iii. Interactions between many genes and all aspects of the environment are involved in forming human intelligence. Genes can’t be changed, but the environment can be altered to affect this very complex phenotype.

台大農藝系 遺傳學 601 20000 chapter 12 slide 1 chapter 13 extensions of mendelian genetic...

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台大農藝系遺傳學 601 20000 chapter 12 slide 1 chapter 13 extensions of mendelian genetic...