GENETICS & EVOLUtion: chromosomal inheritance & mutation

GENETICS & EVOLUtion: chromosomal inheritance & mutation
FISH Karyotype Chapter 7.2

Overview Chromosomal Inheritance Sex-linked Genes
Gene linkage and analysis Mutations Gene Mutations Chromosomal Abberations

Overview: Locating Genes Along Chromosomes
Mendel’s “hereditary factors” were genes, though this wasn’t known at the time Today we can show that genes are located on chromosomes The location of a particular gene can be seen by tagging isolated chromosomes with a fluorescent dye that highlights the gene

Fig. 15-1 Figure 15.1 Where are Mendel’s hereditary factors located in the cell?

Mendelian inheritance has its physical basis in the behavior of chromosomes
Mitosis and meiosis were first described in the late 1800s The chromosome theory of inheritance states: Mendelian genes have specific loci (positions) on chromosomes Chromosomes undergo segregation and independent assortment The behavior of chromosomes during meiosis was said to account for Mendel’s laws of segregation and independent assortment

yellow-round seeds (YyRr)
Fig. 15-2a Yellow-round seeds (YYRR) Green-wrinkled seeds ( yyrr) P Generation y Y r  R R r Y y Meiosis Fertilization y r R Y Figure 15.2 The chromosomal basis of Mendel’s laws Gametes All F1 plants produce yellow-round seeds (YyRr)

0.5 mm All F1 plants produce yellow-round seeds (YyRr) F1 Generation
Fig. 15-2b All F1 plants produce yellow-round seeds (YyRr) 0.5 mm F1 Generation R R y y r r Y Y LAW OF SEGREGATION The two alleles for each gene separate during gamete formation. Meiosis LAW OF INDEPENDENT ASSORTMENT Alleles of genes on nonhomologous chromosomes assort independently during gamete formation. R r r R Metaphase I Y y Y y 1 1 R r r R Anaphase I Y y Y y R r Metaphase II Figure 15.2 The chromosomal basis of Mendel’s laws r R 2 2 Y y Y y y Y Y Y Y y y y Gametes R R r r r r R R 1 4 YR 1 4 yr 1 4 Yr 1 4 yR 3 3

An F1  F1 cross-fertilization
Fig. 15-2c F2 Generation An F1  F1 cross-fertilization 3 3 9 : 3 : 3 : 1 Figure 15.2 The chromosomal basis of Mendel’s laws

Figure 15.2 The chromosomal basis of Mendel’s laws
P Generation Yellow-round seeds (YYRR) Green-wrinkled seeds ( yyrr) Y y  r Y R R r y Meiosis Fertilization R Y y r Gametes All F1 plants produce yellow-round seeds (YyRr) F1 Generation R R y y r r Y Y LAW OF SEGREGATION The two alleles for each gene separate during gamete formation. Meiosis LAW OF INDEPENDENT ASSORTMENT Alleles of genes on nonhomologous chromosomes assort independently during gamete formation. R r r R Metaphase I Y y Y y 1 1 R r r R Anaphase I Y y Y y Figure 15.2 The chromosomal basis of Mendel’s laws R r Metaphase II r R 2 2 Y y Y y y Y Y y Y Y y y Gametes R R r r r r R R 1/4 YR 1/4 yr 1/4 Yr 1/4 yR F2 Generation An F1  F1 cross-fertilization 3 3 9 : 3 : 3 : 1

The Chromosomal Basis of Sex
In humans and other mammals, there are two varieties of sex chromosomes: a larger X chromosome and a smaller Y chromosome Only the ends of the Y chromosome have regions that are homologous with the X chromosome The SRY gene on the Y chromosome codes for the development of testes In humans and some other animals, there is a chromosomal basis of sex determination

Fig. 15-5 X Y Figure 15.5 Human sex chromosomes

Females are XX, and males are XY
Each ovum contains an X chromosome, while a sperm may contain either an X or a Y chromosome Other animals have different methods of sex determination

Figure 15.6 Some chromosomal systems of sex determination
44 + XY 44 + XX Parents 22 + X 22 + Y 22 + X or + Sperm Egg 44 + XX 44 + XY or Zygotes (offspring) (a) The X-Y system 22 + XX 22 + X (b) The X-0 system 76 + ZW 76 + ZZ Figure 15.6 Some chromosomal systems of sex determination (c) The Z-W system 32 (Diploid) 16 (Haploid) (d) The haplo-diploid system

Inheritance of Sex-Linked Genes
The sex chromosomes have genes for many characters unrelated to sex A gene located on either sex chromosome is called a sex-linked gene In humans, sex-linked usually refers to a gene on the larger X chromosomeSex-linked genes follow specific patterns of inheritance For a recessive sex-linked trait to be expressed A female needs two copies of the allele A male needs only one copy of the allele Sex-linked recessive disorders are much more common in males than in females

(a) (b) (c) Sperm Sperm Sperm Eggs Eggs Eggs XNXN  XnY XNXn  XNY
Fig. 15-7 XNXN  XnY XNXn  XNY XNXn  XnY Sperm Xn Y Sperm XN Y Sperm Xn Y Eggs XN XNXn XNY Eggs XN XNXN XNY Eggs XN XNXn XNY XN XNXn XNY Xn XnXN XnY Xn XnXn XnY Figure 15.7 The transmission of sex-linked recessive traits (a) (b) (c)

Some disorders caused by recessive alleles on the X chromosome in humans:
Color blindness Duchenne muscular dystrophy Hemophilia

X Inactivation in Female Mammals
In mammalian females, one of the two X chromosomes in each cell is randomly inactivated during embryonic development The inactive X condenses into a Barr body If a female is heterozygous for a particular gene located on the X chromosome, she will be a mosaic for that character

X chromosomes Allele for orange fur Early embryo: Allele for black fur
Fig. 15-8 X chromosomes Allele for orange fur Early embryo: Allele for black fur Cell division and X chromosome inactivation Two cell populations in adult cat: Active X Inactive X Active X Black fur Orange fur Figure 15.8 X inactivation and the tortoiseshell cat

Linked genes tend to be inherited together
Each chromosome has hundreds or thousands of genes Genes located on the same chromosome that tend to be inherited together are called linked genes Morgan did other experiments with fruit flies to see how linkage affects inheritance of two characters Morgan crossed flies that differed in traits of body color and wing size because they are located near each other on the same chromosome

How Linkage Affects Inheritance
Morgan found that body color and wing size are usually inherited together in specific combinations (parental phenotypes) He noted that these genes do not assort independently, and reasoned that they were on the same chromosome However, nonparental phenotypes were also produced Understanding this result involves exploring genetic recombination, the production of offspring with combinations of traits differing from either parent

b+ vg+ b vg  Parents in testcross b vg b vg b+ vg+ b vg Most or
Fig. 15-UN1 b+ vg+ b vg  Parents in testcross b vg b vg b+ vg+ b vg Most offspring or b vg b vg

EXPERIMENT Fig. 15-9-1 P Generation (homozygous) b+ b+ vg+ vg+
Wild type (gray body, normal wings) Double mutant (black body, vestigial wings)  b+ b+ vg+ vg+ b b vg vg Figure 15.9 How does linkage between two genes affect inheritance of characters?

Wild type (gray body, normal wings) Double mutant (black body, vestigial wings)  b+ b+ vg+ vg+ b b vg vg F1 dihybrid (wild type) Double mutant TESTCROSS  b+ b vg+ vg b b vg vg Figure 15.9 How does linkage between two genes affect inheritance of characters?

Wild type (gray body, normal wings) Double mutant (black body, vestigial wings)  b+ b+ vg+ vg+ b b vg vg F1 dihybrid (wild type) Double mutant TESTCROSS  b+ b vg+ vg b b vg vg Testcross offspring Eggs b+ vg+ b vg b+ vg b vg+ Wild type (gray-normal) Black- vestigial Gray- vestigial Black- normal b vg Figure 15.9 How does linkage between two genes affect inheritance of characters? Sperm b+ b vg+ vg b b vg vg b+ b vg vg b b vg+ vg

EXPERIMENT RESULTS Fig. 15-9-4 P Generation (homozygous) b+ b+ vg+ vg+
Wild type (gray body, normal wings) Double mutant (black body, vestigial wings)  b+ b+ vg+ vg+ b b vg vg F1 dihybrid (wild type) Double mutant TESTCROSS  b+ b vg+ vg b b vg vg Testcross offspring Eggs b+ vg+ b vg b+ vg b vg+ Wild type (gray-normal) Black- vestigial Gray- vestigial Black- normal b vg Figure 15.9 How does linkage between two genes affect inheritance of characters? Sperm b+ b vg+ vg b b vg vg b+ b vg vg b b vg+ vg PREDICTED RATIOS If genes are located on different chromosomes: 1 : 1 : 1 : 1 If genes are located on the same chromosome and parental alleles are always inherited together: 1 : 1 : : RESULTS 965 : 944 : 206 : 185

Genetic Recombination and Linkage
The genetic findings of Mendel and Morgan relate to the chromosomal basis of recombination

Recombination of Unlinked Genes: Independent Assortment of Chromosomes
Mendel observed that combinations of traits in some offspring differ from either parent Offspring with a phenotype matching one of the parental phenotypes are called parental types Offspring with nonparental phenotypes (new combinations of traits) are called recombinant types, or recombinants A 50% frequency of recombination is observed for any two genes on different chromosomes

YR yr Yr yR yr YyRr yyrr Yyrr yyRr Gametes from yellow-round
Fig. 15-UN2 Gametes from yellow-round heterozygous parent (YyRr) YR yr Yr yR Gametes from green- wrinkled homozygous recessive parent ( yyrr) yr YyRr yyrr Yyrr yyRr Parental- type offspring Recombinant offspring

Recombination of Linked Genes: Crossing Over
Morgan discovered that genes can be linked, but the linkage was incomplete, as evident from recombinant phenotypes Morgan proposed that some process must sometimes break the physical connection between genes on the same chromosome That mechanism was the crossing over of homologous chromosomes

Black body, vestigial wings
Fig Testcross parents Gray body, normal wings (F1 dihybrid) Black body, vestigial wings (double mutant) b+ vg+ b vg b vg b vg Replication of chromosomes Replication of chromosomes b+ vg+ b vg b+ vg+ b vg b vg b vg b vg b vg Meiosis I b+ vg+ Meiosis I and II b+ vg b vg+ b vg Meiosis II Recombinant chromosomes Figure Chromosomal basis for recombination of linked genes b+ vg+ b vg b+ vg b vg+ Eggs Testcross offspring 965 Wild type (gray-normal) 944 Black- vestigial 206 Gray- vestigial 185 Black- normal b vg b+ vg+ b vg b+ vg b vg+ b vg b vg b vg b vg Sperm Parental-type offspring Recombinant offspring Recombination frequency 391 recombinants =  100 = 17% 2,300 total offspring

Black body, vestigial wings
Fig a Testcross parents Gray body, normal wings (F1 dihybrid) Black body, vestigial wings (double mutant) b+ vg+ b vg b vg b vg Replication of chromosomes Replication of chromosomes b+ vg+ b vg b+ vg+ b vg b vg b vg b vg b vg Meiosis I b+ vg+ Meiosis I and II b+ vg b vg+ Figure Chromosomal basis for recombination of linked genes b vg Meiosis II Recombinant chromosomes b+ vg+ b vg b+ vg b vg+ b vg Eggs Sperm

965 944 Black- vestigial 206 Gray- vestigial 185 Black- normal
Fig b Recombinant chromosomes b+ vg+ b vg b+ vg b vg+ Eggs Testcross offspring 965 Wild type (gray-normal) 944 Black- vestigial 206 Gray- vestigial 185 Black- normal b vg b+ vg+ b vg b+ vg b vg+ Figure Chromosomal basis for recombination of linked genes b vg b vg b vg b vg Sperm Parental-type offspring Recombinant offspring Recombination frequency 391 recombinants =  100 = 17% 2,300 total offspring

Mapping the Distance Between Genes Using Recombination Data
Alfred Sturtevant, one of Morgan’s students, constructed a genetic map, an ordered list of the genetic loci along a particular chromosome Sturtevant predicted that “the farther apart two genes are, the higher the probability that a crossover will occur between them and therefore the higher the recombination frequency”

A linkage map is a genetic map of a chromosome based on recombination frequencies
Distances between genes can be expressed as map units; one map unit, or centimorgan, represents a 1% recombination frequency Map units indicate relative distance and order, not precise locations of genes

RESULTS Recombination frequencies 9% 9.5% Chromosome 17% b cn vg
Fig RESULTS Recombination frequencies 9% 9.5% Chromosome 17% Figure Constructing a linkage map b cn vg

Genes that are far apart on the same chromosome can have a recombination frequency near 50%
Such genes are physically linked, but genetically unlinked, and behave as if found on different chromosomes Sturtevant used recombination frequencies to make linkage maps of fruit fly genes Using methods like chromosomal banding, geneticists can develop cytogenetic maps of chromosomes Cytogenetic maps indicate the positions of genes with respect to chromosomal features

Mutant phenotypes Short aristae Black body Cinnabar eyes Vestigial
Fig Mutant phenotypes Short aristae Black body Cinnabar eyes Vestigial wings Brown eyes 48.5 57.5 67.0 104.5 Figure A partial genetic (linkage) map of a Drosophila chromosome Long aristae (appendages on head) Gray body Red eyes Normal wings Red eyes Wild-type phenotypes

Abnormal Chromosome Number
Large-scale chromosomal alterations often lead to spontaneous abortions (miscarriages) or cause a variety of developmental disorders In nondisjunction, pairs of homologous chromosomes do not separate normally during meiosis As a result, one gamete receives two of the same type of chromosome, and another gamete receives no copy Alterations of chromosome number or structure cause some genetic disorders

Meiosis I (a) Nondisjunction of homologous
Fig Meiosis I Nondisjunction Figure Meiotic nondisjunction For the Cell Biology Video Nondisjunction in Mitosis, go to Animation and Video Files. (a) Nondisjunction of homologous chromosomes in meiosis I (b) Nondisjunction of sister chromatids in meiosis II

Meiosis I Meiosis II (a) Nondisjunction of homologous
Fig Meiosis I Nondisjunction Meiosis II Nondisjunction Figure Meiotic nondisjunction (a) Nondisjunction of homologous chromosomes in meiosis I (b) Nondisjunction of sister chromatids in meiosis II

Meiosis I Meiosis II Gametes (a) Nondisjunction of homologous
Fig Meiosis I Nondisjunction Meiosis II Nondisjunction Gametes Figure Meiotic nondisjunction n + 1 n + 1 n – 1 n – 1 n + 1 n – 1 n n Number of chromosomes (a) Nondisjunction of homologous chromosomes in meiosis I (b) Nondisjunction of sister chromatids in meiosis II

A monosomic zygote has only one copy of a particular chromosome
Aneuploidy results from the fertilization of gametes in which nondisjunction occurred Offspring with this condition have an abnormal number of a particular chromosome A monosomic zygote has only one copy of a particular chromosome A trisomic zygote has three copies of a particular chromosome Euploid is the correct number of chromosomes in a species. Aneuploid is change in the chromosome number Results from nondisjunction Monosomy - only one of a particular type of chromosome, Trisomy - three of a particular type of chromosome

Polyploidy is common in plants, but not animals
Polyploidy is a condition in which an organism has more than two complete sets of chromosomes Triploidy (3n) is three sets of chromosomes Tetraploidy (4n) is four sets of chromosomes Polyploidy is common in plants, but not animals Polyploids are more normal in appearance than aneuploids

Fig Figure A tetraploid mammal

Mutagens Spontaneous mutations can occur during DNA replication, recombination, or repair Mutagens are physical or chemical agents that can cause mutations Mutations are changes in the genetic material of a cell or virus Gene mutation Chromosomal abberation

Causes of Mutations Spontaneous mutation Induced mutation:
DNA can undergo a chemical change Movement of transposons from one chromosomal location to another Replication Errors 1 in 1,000,000,000 replications DNA polymerase Proofreads new strands Generally corrects errors Induced mutation: Mutagens such as radiation, organic chemicals Many mutagens are also carcinogens (cancer causing) Environmental Mutagens Ultraviolet Radiation Tobacco Smoke

Effect of Mutations on Protein Activity
Point Mutations Involve change in a single DNA nucleotide Changes one codon to a different codon Affects on protein vary: Nonfunctional Reduced functionality Unaffected Frameshift Mutations One or two nucleotides are either inserted or deleted from DNA Protein always rendered nonfunctional Normal : THE CAT ATE THE RAT After deletion: THE ATA TET HER AT After insertion: THE CCA TAT ETH ERA T

Point mutations can affect protein structure and function
Point mutations are chemical changes in just one base pair of a gene The change of a single nucleotide in a DNA template strand can lead to the production of an abnormal protein

Wild-type hemoglobin DNA Mutant hemoglobin DNA 3 C T T 5 3 C A T 5
Fig Wild-type hemoglobin DNA Mutant hemoglobin DNA 3 C T T 5 3 C A T 5 5 G A A 3 5 G T A 3 mRNA mRNA 5 G A A 3 5 G U A 3 Figure The molecular basis of sickle-cell disease: a point mutation Normal hemoglobin Sickle-cell hemoglobin Glu Val

Types of Point Mutations
Point mutations within a gene can be divided into two general categories Base-pair substitutions Base-pair insertions or deletions

Figure 17.23 Types of point mutations
Wild-type DNA template strand 3 5 5 3 mRNA 5 3 Protein Stop Amino end Carboxyl end A instead of G Extra A 3 5 3 5 5 3 5 3 U instead of C Extra U 5 3 5 3 Stop Stop Silent (no effect on amino acid sequence) Frameshift causing immediate nonsense (1 base-pair insertion) T instead of C missing 3 5 3 5 5 3 5 3 A instead of G missing 5 3 5 3 Stop Figure Types of point mutations Missense Frameshift causing extensive missense (1 base-pair deletion) A instead of T missing 3 5 3 5 5 3 5 3 U instead of A missing 5 3 5 3 Stop Stop Nonsense No frameshift, but one amino acid missing (3 base-pair deletion) (a) Base-pair substitution (b) Base-pair insertion or deletion

Silent (no effect on amino acid sequence)
Fig a Wild type DNA template strand 3 5 5 3 mRNA 5 3 Protein Stop Amino end Carboxyl end A instead of G 3 5 5 3 Figure Types of point mutations U instead of C 5 3 Stop Silent (no effect on amino acid sequence)

Wild type DNA template strand 3 5 5 3 mRNA 5 3 Protein Stop
Fig b Wild type DNA template strand 3 5 5 3 mRNA 5 3 Protein Stop Amino end Carboxyl end T instead of C 3 5 5 3 Figure Types of point mutations A instead of G 5 3 Stop Missense

Fig c Wild type DNA template strand 3 5 5 3 mRNA 5 3 Protein Stop Amino end Carboxyl end A instead of T 3 5 5 3 Figure Types of point mutations U instead of A 5 3 Stop Nonsense

Frameshift causing immediate nonsense (1 base-pair insertion)
Fig d Wild type DNA template strand 3 5 5 3 mRNA 5 3 Protein Stop Amino end Carboxyl end Extra A 3 5 5 3 Figure Types of point mutations Extra U 5 3 Stop Frameshift causing immediate nonsense (1 base-pair insertion)

Frameshift causing extensive missense (1 base-pair deletion)
Fig e Wild type DNA template strand 3 5 5 3 mRNA 5 3 Protein Stop Amino end Carboxyl end missing 3 5 5 3 Figure Types of point mutations missing 5 3 Frameshift causing extensive missense (1 base-pair deletion)

Fig f Wild type DNA template strand 3 5 5 3 mRNA 5 3 Protein Stop Amino end Carboxyl end missing 3 5 5 3 Figure Types of point mutations missing 5 3 Stop No frameshift, but one amino acid missing (3 base-pair deletion)

Substitutions A base-pair substitution replaces one nucleotide and its partner with another pair of nucleotides Silent mutations have no effect on the amino acid produced by a codon because of redundancy in the genetic code Missense mutations still code for an amino acid, but not necessarily the right amino acid Nonsense mutations change an amino acid codon into a stop codon, nearly always leading to a nonfunctional protein

Insertions and Deletions
Insertions and deletions are additions or losses of nucleotide pairs in a gene These mutations have a disastrous effect on the resulting protein more often than substitutions do Insertion or deletion of nucleotides may alter the reading frame, producing a frameshift mutation

Alterations of Chromosome Structure
Breakage of a chromosome can lead to four types of changes in chromosome structure: Deletion removes a chromosomal segment Duplication repeats a segment Inversion reverses a segment within a chromosome Translocation moves a segment from one chromosome to another

Reciprocal translocation
Fig A B C D E F G H A B C E F G H Deletion (a) A B C D E F G H A B C B C D E F G H Duplication (b) A B C D E F G H A D C B E F G H (c) Inversion Figure Alterations of chromosome structure A B C D E F G H M N O C D E F G H (d) Reciprocal translocation M N O P Q R A B P Q R

Human Disorders Due to Chromosomal Alterations
Alterations of chromosome number and structure are associated with some serious disorders Some types of aneuploidy appear to upset the genetic balance less than others, resulting in individuals surviving to birth and beyond These surviving individuals have a set of symptoms, or syndrome, characteristic of the type of aneuploidy

Down Syndrome (Trisomy 21)
Down syndrome is an aneuploid condition that results from three copies of chromosome 21 It affects about one out of every 700 children born in the United States The frequency of Down syndrome increases with the age of the mother, a correlation that has not been explained short stature eyelid fold flat face stubby finger wide gap between first and second toes

Fig Figure Down syndrome

Aneuploidy of Sex Chromosomes
Nondisjunction of sex chromosomes produces a variety of aneuploid conditions Klinefelter syndrome is the result of an extra chromosome in a male, producing XXY individuals Monosomy X, called Turner syndrome, produces X0 females, who are sterile; it is the only known viable monosomy in humans Turner syndrome (XO) Female with single X chromosome Short, with broad chest and widely spaced nipples Can be of normal intelligence and function with hormone therapy Klinefelter syndrome (XXY) – a male Male with underdeveloped testes and prostate; some breast overdevelopment Long arms and legs; large hands Near normal intelligence unless XXXY, XXXXY, etc. No matter how many X chromosomes, presence of Y renders individual male

Changes in Sex Chromosome
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. a. Turner syndrome b. Klinefelter syndrome a: Courtesy UNC Medical Illustration and Photography; b: Courtesy Stefan D. Schwarz,

Disorders Caused by Structurally Altered Chromosomes
The syndrome cri du chat (“cry of the cat”), results from a specific deletion in chromosome 5 A child born with this syndrome is mentally retarded and has a catlike cry; individuals usually die in infancy or early childhood Certain cancers, including chronic myelogenous leukemia (CML), are caused by translocations of chromosomes

Translocated chromosome 22 (Philadelphia chromosome)
Fig Reciprocal translocation Normal chromosome 9 Translocated chromosome 9 Translocated chromosome 22 (Philadelphia chromosome) Normal chromosome 22 Figure Translocation associated with chronic myelogenous leukemia (CML)

Carcinogenesis Development of cancer involves a series of mutations
Proto-oncogenes – Stimulate cell cycle Tumor suppressor genes – inhibit cell cycle Mutation in oncogene and tumor suppressor gene: Stimulates cell cycle uncontrollably Leads to tumor formation

Cell Signaling Pathway
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. inhibiting growth factor receptor plasma membrane signal transducers cytoplasm transcription factor protein that is unable to inhibit the cell cycle or promote apoptosis nucleus mutated tumor suppressor gene Cell signaling pathway that stimulates a mutated tumor suppressor gene

Cell Signaling Pathway
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. stimulating growth factor receptor plasma membrane signal transducers cytoplasm transcription factor protein that overstimulates the cell cycle nucleus oncogene Cell signaling pathway that stimulates a proto-oncogene

Some inheritance patterns are exceptions to the standard chromosome theory
There are two normal exceptions to Mendelian genetics One exception involves genes located in the nucleus, and the other exception involves genes located outside the nucleus

Genomic Imprinting For a few mammalian traits, the phenotype depends on which parent passed along the alleles for those traits Such variation in phenotype is called genomic imprinting Genomic imprinting involves the silencing of certain genes that are “stamped” with an imprint during gamete production

Normal Igf2 allele is expressed Paternal chromosome Maternal
Fig a Normal Igf2 allele is expressed Paternal chromosome Maternal chromosome Normal Igf2 allele is not expressed Wild-type mouse (normal size) Figure Genomic imprinting of the mouse Igf2 gene (a) Homozygote

Mutant Igf2 allele Mutant Igf2 allele inherited from mother
Fig b Mutant Igf2 allele inherited from mother Mutant Igf2 allele inherited from father Normal size mouse (wild type) Dwarf mouse (mutant) Normal Igf2 allele is expressed Mutant Igf2 allele is expressed Figure Genomic imprinting of the mouse Igf2 gene Mutant Igf2 allele is not expressed Normal Igf2 allele is not expressed (b) Heterozygotes

Normal Igf2 allele is expressed Paternal chromosome Maternal
Fig Normal Igf2 allele is expressed Paternal chromosome Maternal chromosome Normal Igf2 allele is not expressed Wild-type mouse (normal size) (a) Homozygote Mutant Igf2 allele inherited from mother Mutant Igf2 allele inherited from father Normal size mouse (wild type) Dwarf mouse (mutant) Figure Genomic imprinting of the mouse Igf2 gene Normal Igf2 allele is expressed Mutant Igf2 allele is expressed Mutant Igf2 allele is not expressed Normal Igf2 allele is not expressed (b) Heterozygotes

It appears that imprinting is the result of the methylation (addition of –CH3) of DNA
Genomic imprinting is thought to affect only a small fraction of mammalian genes Most imprinted genes are critical for embryonic development

Fig. 15-UN3

Inheritance of Organelle Genes
Extranuclear genes (or cytoplasmic genes) are genes found in organelles in the cytoplasm Mitochondria, chloroplasts, and other plant plastids carry small circular DNA molecules Extranuclear genes are inherited maternally because the zygote’s cytoplasm comes from the egg The first evidence of extranuclear genes came from studies on the inheritance of yellow or white patches on leaves of an otherwise green plant

Some defects in mitochondrial genes prevent cells from making enough ATP and result in diseases that affect the muscular and nervous systems For example, mitochondrial myopathy and Leber’s hereditary optic neuropathy

You should now be able to:
Explain the chromosomal theory of inheritance and its discovery Explain why sex-linked diseases are more common in human males than females Distinguish between sex-linked genes and linked genes Explain how meiosis accounts for recombinant phenotypes Explain how linkage maps are constructed

Explain how nondisjunction can lead to aneuploidy
Define trisomy, triploidy, and polyploidy Define mutation Distinguish between different gene mutations Distinguish among deletions, duplications, inversions, and translocations Explain genomic imprinting Explain why extranuclear genes are not inherited in a Mendelian fashion

Morgan’s Experimental Evidence Imagine that Morgan had used a grasshopper (2N = 24 and an XX, XO sex determination system). Predict where the first mutant would have been discovered. on the O chromosome of a male on the X chromosome of a male on the X chromosome of a female on the Y chromosome of a male Answer: there are a variety of sex determination mechanisms and to help students understand the significance of Morgan’s work. Answer b. Related to material on textbook page 288 and Figure 15.6. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.

The Chromosomal Basis of Sex Think about bees and ants, groups in which males are haploid. Which of the following are accurate statements about bee and ant males when they are compared to species in which males are XY and diploid for the autosomes? Bee males have half the DNA of bee females whereas human males have nearly the same amount of DNA that human females have. Considered across the genome, harmful (deleterious) recessives will negatively affect bee males more than Drosophila males. Human and Drosophila males have sons but bee males do not. Inheritance in bees is like inheritance of sex-linked characteristics in humans. none of the above Answer: there are a variety of sex determination mechanisms among different species. Answers a, b, and c are correct. Answer d is incorrect because male bees have no sons. Related to material in textbook Figure 15.6 and pages Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.

The Chromosomal Basis of Sex In some Drosophila species there are genes on the Y chromosome that do not occur on the X chromosome. Imagine that a mutation of one gene on the Y chromosome reduces the size by half of individuals with the mutation. Which of the following statements is accurate with regard to this situation? This mutation occurs in all offspring of a male with the mutation. This mutation occurs in all male but no female offspring of a male with the mutation. This mutation occurs in all offspring of a female with the mutation. This mutation occurs in all male but no female offspring of a female with the mutation. This mutation occurs in all offspring of both males and females with the mutation. Answer: understand sex-linkage by looking at the effect of genes on the Y chromosome. Answer b is correct because all the sons of the mutant male receive his Y chromosome. Related to material on textbook page 290. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.

The Chromosomal Basis of Sex Imagine that a deleterious recessive allele occurs on the W chromosome of a chicken (2N = 78). Where would it be most likely to appear first in a genetics experiment? in a male because there is no possibility of the presence of a normal, dominant allele in a male because it is haploid in a female because there is no possibility of the presence of a normal, dominant allele in a female because all alleles on the W chromosomes are dominant to those on the Z chromosome none of the above Answer: This should be a very easy question if students have studied Figure Although it is easy, it should reinforce the idea that there are different methods of sex determination. The correct answer is e Related to material on textbook page 290. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.

Inheritance of Sex-Linked Genes In cats, a sex-linked gene affects coat color. The O allele produces an enzyme that converts eumelanin, a black or brown pigment, into phaeomelanin, an orange pigment. The o allele is recessive to O and produces a defective enzyme, one that does not convert eumelanin into phaeomelanin. Which of the following statements is/are accurate? The phenotype of O-Y males is orange because the functional allele O converts eumelanin into phaeomelanin. The phenotype of o-Y males is black/brown because the non-functional allele o does not convert eumelanin into phaeomelanin. The phenotype of OO and Oo males is orange because the functional allele O converts eumelanin into phaeomelanin. The phenotype of Oo males is mixed orange and black/brown because the functional allele O converts eumelanin into phaeomelanin in some cell groups (orange) and because in other cell groups the non-functional allele o does not convert eumelanin into phaeomelanin. The phenotype of O-Y males is orange because the non-functional allele O does not convert eumelanin into phaeomelanin while the phenotype of o-Y males is black/brown because the functional allele o converts eumelanin into phaeomelanin. Answer: This is the first of two questions on the sex-linked trait of orange coat in cats. It focuses on the color of males and the action of the enzyme that converts eumelanin (brown/black pigment) to phaeomelanin (orange pigment). Male genotypes will be either O-Y or o-Y, with phenotypes of either orange or black/brown, respectively. In O-Y males, the eumelanin is converted to phaeomelanin and in o-Y males, the eumelanin is unchanged. To answer this question, males have only one copy of the gene and understand that a functional allele produces an enzyme that catalyzes the chemical reaction. Answers a and b are correct because they correctly show male genotypes and their associated phenotypes. Related to material on textbook pp and Figure 15.8. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.

Inheritance of Sex-Linked Genes In cats, a sex-linked gene affects coat color. The O allele produces an enzyme that converts eumelanin, a black or brown pigment, into phaeomelanin, an orange pigment. The o allele is recessive to O and produces a defective enzyme, one that does not convert eumelanin into phaeomelanin. Which of the following statements is/are accurate? The phenotype of O-Y females is orange because the functional allele O converts eumelanin into phaeomelanin. The phenotype of o-Y females is black/brown because the non-functional allele o does not convert eumelanin into phaeomelanin. The phenotype of OO and Oo females is orange because the functional allele O converts eumelanin into phaeomelanin. The phenotype of Oo females is mixed orange and black/brown because the functional allele O converts eumelanin into phaeomelanin in some cell groups (orange) and because in other cell groups the non-functional allele o does not convert eumelanin into phaeomelanin. The phenotype of O-Y females is orange because the non-functional allele O does not convert eumelanin into phaeomelanin while the phenotype of o-Y males is black/brown because the functional allele o converts eumelanin into phaeomelanin. Answer: This is the second of two questions on the sex-linked trait of orange coat in cats. It focuses on the color of females, the action of the enzyme that converts eumelanin (brown/black pigment) to phaeomelanin (orange pigment), and the development of the Barr body. Female genotypes will be OO (phenotype orange), oo (phenotype black/brown), or Oo (mixed orange and black/brown). In OO females, the eumelanin is converted to phaeomelanin, and in oo females, the eumelanin is unchanged. In Oo females, hairs produced by cells in which the inactivated X chromosome has the O allele are black/brown because the active allele o does not affect the eumelanin. In areas of the Oo females where the cells contain inactivated X chromosomes with the o allele, the hairs are orange because the enzyme produces phaeomelanin. To answer this question 1) know that females have two copies of the gene, 2) understand that a functional allele produces an enzyme that catalyzes a chemical reaction, and 3) realize that the development of a Barr body results in patches of different colored fur. Answer d is correct. Related to material on textbook pages and Figure 15.8. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.

X Inactivation in Female Mammals Imagine two species of mammals that differ in the timing of Barr body formation during development. Both species have genes that determine coat color, O for the dominant orange fur and o for the recessive black/brown fur, on the X chromosome. In species A, the Barr body forms during week 1 of a 6-month pregnancy whereas in species B, the Barr body forms during week 3 of a 5-month pregnancy. What would you predict about the coloration of heterozygous females (Oo) in the two species? Both species will have similar sized patches of orange and black/brown fur. Species A will have smaller patches of orange or black/brown fur than will species B. The females of both species will show the dominant fur color, orange. Answer: This question relates genetics, embryonic development, and Barr body formation. The sizes of the patches should be related to the timing of Barr body formation in development. Early formation of the Barr body would mean that larger clusters of cells are affected by which X chromosome is inactivated compared to later Barr body formation. Thus, answer b is the correct answer. Related to material on textbook pages Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.

Mapping the Distance between Genes Imagine a species with three loci thought to be on the same chromosome. The recombination rate between locus A and locus B is 35% and the recombination rate between locus B and locus C is 33%. Predict the recombination rate between A and C. The recombination rate between locus A and locus C is either 2% or 68%. The recombination rate between locus A and locus C is probably 2%. The recombination rate between locus A and locus C is either 2% or 50%. The recombination rate between locus A and locus C is either 2% or 39%. The recombination rate between locus A and locus C cannot be predicted. Answer: The recombination rate between loci A and B is 35%. Locus C can be either between A and B or on the opposite side of B from A. If locus C is in between locus A and locus B (ACB), the distance between locus A and C would be 2%. Locus C cannot be on the other side of A from B (CAB) because the recombination rate would have to be higher than 35%. Thus far, answers a, b, c, and d could be correct. If locus C is on the other side of locus B from locus A (ABC), adding the two recombination rates gives a prediction of 68%, but the maximum recombination rate between two loci is 50%, the same frequency of recombinants that are on different chromosomes. Using this information, answer a cannot be correct because a recombination rate cannot be greater than 50%. Answer b could be right but it is not the only possible placement and so answer b is incomplete. Answer c is the best answer (2% if C is between A and B, and 50% if C is on the other side of B than A). Related to material on textbook pages Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.

Triploid species are usually sterile (unable to reproduce) whereas tetraploids are often fertile. Which of the following are likely good explanations of these facts? In mitosis, some chromosomes in triploids have no partner at synapsis, but chromosomes in tetraploids do have partners. In meiosis, some chromosomes in triploids have no partner at synapsis, but chromosomes in tetraploids do have partners. In mitosis, some chromosomes in tetraploids have no partner at synapsis, but chromosomes in triploids do have partners. In meiosis, some chromosomes in tetraploids have no partner at synapsis, but chromosomes in triploids do have partners. Answer: The point of this question is to think about mitosis and meiosis (Chapter 13) in relation to polyploids. To answer this question, draw chromosomes of a triploid and a tetraploid as they go through mitosis and meiosis. Answer b is correct—one-third of the chromosomes do not have a partner. Related to material on textbook page 297. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.

Chromosomal rearrangements can occur after chromosomes break
Chromosomal rearrangements can occur after chromosomes break. Which of the following statements are most accurate with respect to alterations in chromosome structure? Chromosomal rearrangements are more likely to occur in mammals than in other vertebrates. Translocations and inversions are not deleterious because no genes are lost in the organism. Chromosomal rearrangements are more likely to occur during mitosis than during meiosis. An individual that is homozygous for a deletion of a certain gene is likely to be more damaged than is one that is homozygous for a duplication of that same gene because loss of a function can be lethal. Answer: Chromosomal rearrangements are important in evolution. For example, duplications provide raw material on which natural selection can act (e.g., the globin genes are thought to have arisen via gene duplication). This question will make you think about the consequences of chromosomal rearrangements. Answer d is correct. Related to material on textbook page 298. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.

Imagine that you could create medical policy for a country
Imagine that you could create medical policy for a country. In this country it is known that the frequency of Down syndrome babies increases with increasing age of the mother and that the severity of characteristics varies enormously and unpredictably among affected individuals. Furthermore, financial resources are severely limited, both for testing of pregnant women and for supplemental training of Down syndrome children. The graph on the next slide shows the incidence of Down syndrome as a function of maternal age. Which of the following policies would you implement? No testing of pregnant women should be conducted and all the health care money should be used for training of Down syndrome children. The health care system should provide testing only for women over 30. The health care system should provide testing only for women over 40. The health care system should require termination of all Down syndrome fetuses. The health care system should provide training for the 30% most seriously affected children only. Answer: There is no correct answer to this question. It will be important to emphasize that financial resources are limited. We often act as if such funds are, or should be, unlimited. A variety of arguments could be made for and against all of the suggested policies. Medical testing is likely to be much cheaper than training, meaning that answer a is not really a realistic policy. On the other hand, fewer than 4% of fetuses are affected in the highest risk group, so perhaps no testing is needed. Answers b and c look at “break points” in the graph—ages when Down syndrome incidence increases somewhat sharply. Answer d will be disliked by many people, partially on the basis that no fetuses should be aborted and partially on the basis that some Down syndrome people are not severely affected and can lead productive lives. Answer e contains a purely arbitrary number (30%) and it could be argued that this is the only practical policy. The graph is from This site contains a lot of information. Related to information on textbook page 299. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.

The lawyer for a defendant in a paternity suit asked for DNA testing of a baby girl. Which of the following set of results would demonstrate that the purported father was not actually the genetic father of the child? The mitochondrial DNA of the child and “father” did not match. DNA sequencing of chromosome #5 of the child and “father” did not match. The mitochondrial DNA of the child and “mother” did not match. DNA sequencing of chromosome #5 of the child and “mother” did not match. The mitochondrial DNA of the child and “father” matched but the mitochondrial DNA of the child and “mother” did not. Answer: The purpose of this question is to understand that mitochondria are inherited from the mother. Mitochondrial DNA between a mother and child should match but there is no expectation of a match between father and child. Autosomal DNA should match both parents. Therefore, only answer b would demonstrate that the man was not the father of the child in question. Related to material on textbook pages Copyright © 2008 Pearson Education, Inc., publishing as Pearson Benjamin Cummings.

GENETICS & EVOLUtion: chromosomal inheritance & mutation

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GENETICS & EVOLUtion: chromosomal inheritance & mutation

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