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Chapter 13 Extensions of and Deviations from Mendelian Genetic Principles Copyright © 2010 Pearson Education Inc.

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Presentation on theme: "Chapter 13 Extensions of and Deviations from Mendelian Genetic Principles Copyright © 2010 Pearson Education Inc."— Presentation transcript:

1 Chapter 13 Extensions of and Deviations from Mendelian Genetic Principles Copyright © 2010 Pearson Education Inc.

2  Not all genes have only two forms (alleles); many have multiple alleles. ◦ a.Differences in the DNA sequence of a gene result in multiple alleles. ◦ b.No matter how many alleles for the gene exist in the multiple allelic series, a diploid individual will have only two alleles, one on each homologous chromosome.

3  The number of possible genotypes in a multiple allelic series depends on how many alleles are involved ◦ a.The formula n(n1+1)/2 calculates possible genotypes for n alleles. ◦ b.Of the genotypes predicted by the formula, n are homozygotes, and n(n 2-1)/2 are heterozygotes.

4  ABO blood groups result from a series of three alleles (I A, I B, and i) that combine to produce four phenotypes (A, B, AB, and O).  Both I A and I B are dominant to i, while I A and I B are codominant to each other. The resulting phenotypes are: ◦ a.People with genotype i/i are blood type O. ◦ b.People with genotype I A /I A or I A /i are blood type A. ◦ c.People with genotype I B /I B or I B /i are blood type B. ◦ d.People with genotype I A /I B are blood type AB.

5  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.I A /i and i/i (one type A, the other type O). ◦ c.I A /i and I A /i (both type A). ◦ d.I B /i and i/i (one type B, the other type O). ◦ e.I B /i and I B /i (both type B). ◦ f.I A /i and I B /i (one type A, the other type B).  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 (I A /I B ) 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.

6  Cellular antigens are important in blood transfusions, since recipient antibodies may respond to antigens on donor cells. Recipients generally can accept blood containing antigens the recipient already has.. ◦ a.Type A blood can be transfused into type A or AB recipients. ◦ b.Type B blood can be transfused into type B or AB recipients. ◦ c.Type AB blood can be transfused only into type AB recipients, the only ones who will recognize both antigens as self. ◦ d.Type O blood has neither antigen, and so can be transfused into a recipient of any blood type.

7  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. ◦ b.In most people, the glycolipid is the H antigen.  i. The I A gene product is a glycosyltransferase that adds a-N- acetylgalactosamine to the H antigen, converting it to the A antigen.  ii. The I B gene product is a glycosyltransferase that adds galactose to its polysaccharide, converting the H antigen to the B antigen.  iii. Both enzymes are present in an I A /I B 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.

8  Drosophila has over 100 mutant alleles at the eye- color locus on the X chromosome. Homozygotes for each allele have eyes of a distinct color in the spectrum between white and red, depending on how much pigment depositing function remains in the encoded protein.

9  The eosin allele (w e ) is recessive to wild type.  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 (w e /w e ) female with a white-eyed male (w/Y), the F 1 females are all w e /w. They have eosin eyes, showing that w e is dominant over w.

10  Next the eosin-eyed F 1 females (w e /w) are crossed with red- eyed males w + /Y. ◦ i.All female progeny are red-eyed (w + /w or w + /w e ). ◦ ii.Male progeny are 1 ⁄ 2 eosin-eyed (w e /Y), and 1 ⁄ 2 white-eyed (w/Y).

11  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.  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.

12  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.

13  Incomplete dominance is an allelic relationship where dominance is only partial.  In a heterozygote, the recessive allele is not expressed. The dominant allele is unable to produce the full phenotype as in homozygous dominant individual. The result is intermediate phenotype.  Palomino horses when interbred, the progeny are: ◦ a. 1 ⁄ 4 cremello (cream colored) with genotype C cr /C cr. ◦ b. 1 ⁄ 2 palomino with genotype C/C cr. ◦ c. 1 ⁄ 4 light chestnut (full coat color) with genotype C/C.

14  Incomplete dominance often occurs in plants. An example is flower color in snapdragons involving two alleles, C R and C W. Red-flowered plants (C R /C R ) crossed with white-flowered ones (C W /C W ) produce all pink progeny (C R /C W ).  The sickle-cell mutation in humans is another example.

15  In codominance, the heterozygote’s phenotype includes the phenotypes of both homozygotes. Examples include: ◦ a.The ABO blood series, in which a heterozygous I A /I B 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 L M /L M.  ii.Type MN, with genotype L M /L N.  iii.Type N, with genotype L N /L N.

16  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. ◦ d.Gain-of-function alleles may also show incomplete dominance, with heterozygotes intermediate in phenotype between wild-type and fully mutant homozygotes.

17  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.

18  Yellow crossed with nonyellow results in a ratio of 1 yellow : 1 nonyellow. This suggests yellow is heterozygous.  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).  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.  Yellow is an allele of the agouti locus, designated A Y. shows the yellow X yellow cross. ◦ i. The cross is A Y /A 3 A Y /A, and death of the homozygous yellow animals (A Y /A Y ) 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.

19  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.  Heterozygous mice (A Y /A) express the A Y allele at high levels in all tissues during all developmental stages. Tissue- specific regulation appears to be lost in the A Y allele.  The A Y allele transcript RNA is 50% longer than that of the wild-type allele (A 1 ). This is because: ◦ (1)The A Y 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.  Embryonic lethality of A Y /A Y mice probably results from lack of Raly gene activity rather than from the defective agouti gene.

20  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.  Hemophilia, resulting from an X-linked recessive allele, is lethal if untreated.  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 lethal are rare, since death before reproduction would eliminate the gene from the pool.

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

22  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. ◦ 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.

23  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.

24  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.

25  An example is neurofibromatosis. ◦ 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.  Incomplete penetrance and variable expressivity complicate medical genetics and genetic counseling.

26  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.  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.

27  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.  (4) This gene shows variable expressivity in:  (a)Age of onset.  (b)Site of baldness (crown or forehead).  (c)Degree of hair loss.

28  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 Siamese cats. ◦ a.Siamese phenotype occurs in cats homozygous for the recessive c S allele of the C (albino) locus, which encodes a tyrosinase. ◦ b.Tyrosinase is required in the melanin synthesis pathway. Activity of the tyrosinase encoded by c S is temperature sensitive.  i. Kittens are uniformly warm, and so stay light colored.  ii. As cats grow, extremities become cooler, and tyrosinase becomes active, allowing melanin to be made and fur on the points to become darker.  iii. Color on the points depends on other coat color genes in the animal.

29  Chemicals can have significant effects. Example: ◦ a.Phenylketonuria (PKU) is an autosomal recessive defect in metabolism of the amino acid phenylalanine. If the defect is not treated by restricting phenylalanine in the diet, severe mental retardation and other symptoms result.

30  Some maternally derived phenotypes are produced by the maternal nuclear genome (maternal effect) rather than inherited as extranuclear genes (maternal inheritance). ◦ a.Proteins and/or mRNA deposited in the oocyte before fertilization direct early development in the embryo. ◦ b.The genes encoding these products are on nuclear chromosomes. No mtDNA is involved.

31  Shell coiling in the snail Limnaea peregra. ◦ a.Shell coiling is determined by a pair of nuclear alleles, with the dominant D allele producing a dextral (right) coil and the recessive d producing sinistral (left) coiling. ◦ b.The shell-coiling phenotype is always determined by the mother’s genotype. ◦ c.In all crosses of true-breeding dextral and sinistral snails, the F 1 s have the same genotype (D/d) but the reciprocal crosses produce different phenotypes.

32  A dextral female (D/D) crossed with a sinistral (d/d) male produces a dextral F 1 (D/d). ◦ (1) The F 2 genotypes have a 1:2:1 ratio (D/D : D/d : d/d). All F 2 snails, including those with genotype d/d, have dextral shells. ◦ (2) Selfing the F 2 produces an F 3 that is 3 ⁄ 4 dextral and 1 ⁄ 4 sinistral. The sinistral snails are the progeny of F 2 d/d mothers (who had dextral shells).

33  A sinistral female (d/d) crossed with a dextral male (D/D) produces a sinistral F 1 (D/d). ◦ (1) The F 2 genotypes also have a 1:2:1 ratio (D/D : D/d : d/d). All F 2 snails have sinistral shells. ◦ (2) Selfing the F 2 produces an F 3 that is all dextral, due to the D/d genotype of the F 2 mothers (who had sinistral shells).

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