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FBY 0416 - Chapter 4 - Genetic Inheritance (Latest).pptx

Here are the possible genotypes of the gametes from this individual: RH Rh rH rh The individual is heterozygous for both genes, so it can produce 4 different types of gametes by the principle of independent assortment. The gametes contain either the R or r allele randomly combined with either the H or h allele.

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CHAPTER 4 - GENETIC INHERITANCE
At the end of the topics students should able to: 1. Define terminologies used in genetic inheritance. 2. State Mendel’s first law (law of segregation) and second law (law of independent assortment). 3. Calculate genotypic and phenotypic ratio. 4. Explain codominant alleles, incomplete dominant alleles, multiple alleles, linked genes and sex-linked genes. 5. Calculate genetic distance (map unit).
Genetic inheritance Terms & concepts Mendel’s experiment Pisum sativum – garden pea Monohybrid crosses Dihybrid crosses Law of segregation Law of independent assortment Deviation from Mendel’s Law Codominance Incomplete dominance Multiple allele Polygene Lethal allele Linked gene Genetic mapping Pedigree analysis
What is a gene? A. A factor that controls a heritable characteristic B. Something on a chromosome C. Information stored in a segment of DNA D. Something that encodes a protein
▪ Inheritance is how traits, or characteristics, are passed on from generation to generation. ▪ Chromosomes are made up of genes, which are made up of DNA. ▪ Genetic material (genes, chromosomes, DNA) is found inside the nucleus of acell. ▪ Particulate Hypothesis of Inheritance. ▪ Parents pass on to their offspring separate and distinct factors (today called genes) that are responsible for inherited traits.
▪ Gene - A piece of DNA that encodes a particular trait. EX a gene for eye-color ▪ Allele-an alternate form of a gene.  E.g. - allele for blue eyes and allele for brown eyes ▪ Locus — the location of a gene on a chromosome. Plural =loci ▪ Hybrid -The offspring of genetically dissimilarfrom parents. ▪ Phenotype-the physical appearance of an individual, determined by his or her genotype. (black, brown, short, tall, etc). ▪ Genotype-the genetic composition of an individual or combination of alleles. (BB, Bb, bb)
 Homozygous - containing a pair of the same alleles.Can be ▪ Homozygous recessive - two recessive alleles e.g. bb, or ▪ Homozygous dominant – two dominant alleles e.g. BB ▪ Heterozygous - containing two different alleles. E.g. Bb ▪ Carrier – an individual who has a recessive allele of a gene that does not have an effect on their phenotype. ▪ Dominant allele — expressed whether alone or in pairs. Symbolized by a capital letter.  E.g. Brown eye allele=B ▪ Recessive allele — expressed only in the absence of a dominant allele. Symbolized by lowercase.  E.g. blue eye allele=b
▪ Austrian Monk. ▪ Is considered “The Father of Genetics" ▪ Experimented with “pea plants”. ▪ Used pea plants because they:  were available  reproduced quickly  showed obvious differences in the traits Understood that there was something that carried traits from one generation to the next- “FACTOR”.
1. Law of Dominance: - In a cross of parents that are pure for contrasting traits, only one form of the trait will appear in the next generation. - Offspring that are hybrid for a trait will have only the dominant trait in the phenotype. What happens when the F1’s are crossed?
The F1 crossed produced the F2 generation and the lost trait appeared with predictable ratios. This led to the formulation of the current model of inheritance. ● In a hybrid union, the allele which expresses itself phenotypically = dominant allele while the other allele which fails to express itself phenotypically = recessive allele. ● The hybrid individual shows phenotypically only the dominant character. ● The law of dominance is often described as Mendel’s first law of inheritance.
2. Law of Segregations: -During the formation of gametes (eggs or sperm), the two alleles (hereditary units) responsible for a trait separate from each other. -Alleles for a trait are then "recombined" at fertilization, producing the genotype for the traits of the offspring. 3. Law of IndependentAssortment: - Alleles for different traits are distributed to sex cells(& offspring) independently of one another.
▪A hybrid produced by crossing t wo individuals with different alleles at one genetic locus of interest. ▪In Mendel's experiments with garden peas (Pisum sativum) . ▪ Earliest and simplest breeding experiment. ▪ Organisms showed contrasting traits in character were crossed.
1. they were easy to grow 2. they had a short life cycle 3. their pollination could be controlled 4. they have easily observable characteristics.
He studied 7 characteristics, each of which has two contrasting alternatives. seed shape : round / wrinkled pod color : yellow / green seed color : yellow / green flower color : purple / white pod shape : inflated / constricted flower position : axial / terminal plant height : tall / dwarf
Totest the particulate hypothesis, Mendel crossed true-breeding plants that had two distinct and contrasting and found that the resulting offspring ALWAYSlooked like just one of the parents, not a combination of the two. What is meant by “true breeding?”
True breed organism– homozygous dominant or homozygous recessive. ▪ A true breeding organism or a purebred, is an organism that always •2 true-breeding purple flower(PP) is crossed with white flower(pp). •When self-pollinate or crossed among themselves – all offspring produced will be identical to the parent which all have purple flower.
▪ This 1st crossing is calledhybridisation ▪ True breeding parents are called – P generation ▪ Their offspring are called – F1 generation or (First Filial Generation) ▪ Next he allowed the F1 generation to self-pollinate ▪ This will produce the Second Filial Generation orF2 generation. What happened ??
Conclusion: Recessive trait had not been destroyed or deleted in F1 generation, but merely masked by dominant trait.
▪ 4 concepts: 1. Alternative versions of genes account for variations in inherited characters.  Gene for flower colour exists in 2 versions: ▪ Purple and white i.e. P and p. ▪ These alternative versions of a gene are called Alleles. ▪ Location of a gene on a chromosome is called Locus.
Alleles: alternative versions of a gene. The gene for a particular inherited character resides at a specific locus (position) on homologous chromosome. For each character, an organism inherits two alleles, one from each parent.
2.  The 2 alleles at a particular locus may be identical (true-breeding plants,TT or tt) or it may differ as in the F1 hybrids(Tt) 3. If the 2 alleles at a locus differ, then one, the dominant allele determines the organism’s appearance, the other, the recessive allele, has no noticeable effect on the organism. 4. The 2 alleles for a heritable character segregate (separate) during gamete formation and end up in different gametes. This is the Law of Segregation orThe 1st Mendel’sLaw.
▪ Two alleles for a gene segregate during gamete formation and are rejoined at random, one from each parent, during fertilization.
▪ Used to determine the outcome of a cross between two individuals.
How does a genotype ratio differ from the phenotype ratio?
When any individual produces gametes, the alleles separate, so that each gamete receives only one member of the pair of alleles. ▪ An exception to this rule is linked genes Gametes:
Is designed to reveal whether an organism that displays the dominant phenotype is homozygous or heterozygous. A cross between a recessive homozygous and an organism of dominant phenotype, but unknown genotype.  If an organism displays a dominant characteristic, it may possess 2 dominant alleles (homologous) or a dominant and recessive allele for that characteristic (heterozygous)  To find out which in the case, the organism is crossed with one displaying the recessive characteristic.  If all the offspring show the dominant characteristic then the organism is homozygous, but if half show the recessive characteristic, then the organism is heterozygous.
 A mating between individuals of the parental generation (P) and the first generation (F1). ▪ A backcross can be made to the dominant parental type or to the recessive parental type. ▪ If the backcross is to the dominant parent, all offspring show the dominant phenotype. ▪ If the backcross is to the recessive parent (a testcross), ½ the offspring have the dominant phenotype and ½ have the recessive phenotype.
▪ A back cross=test cross if a parent is homozygous recessive.
 A cross reversing the roles of males and females to confirm the results obtained from an earlier cross. @  A cross with the phenotype of each sex reversed as compared with the original cross.
▪ Each pair of alleles segregates independently during gamete formation.  Allele for one gene will be found within a resulting gamete independently of the allele for a different gene in the same gamete. ▪ Mendel experimented this using 2 pairs of contrasting traits at a time i.e. dihybrid cross. ▪ This law applies only to genes (allele pairs) located on different chromosomes. ▪ Genes located near each other or in the same chromosome tend to be inherited together and have more complex inheritance patterns.
▪ Each pair of alleles segregates independently during gamete formation.
It applies to genes that lie on separate chromosomes. It does not apply when genes lie on the same chromosome.
▪ The inheritance of two separate traits in a single cross ▪ for example: RRYY x rryy Let: R represent round seed r represent wrinkled seed Y represent yellow seed y represent green seed Seed shape Seed colour
Parental Phenotypes: Wrinkled, seeds Parental genotypes: rryy Round, yellow seeds RRYY x x green R R Y Y r r y y Gametes: R Y F1 genotypes: F1 phenotypes: 100% round, yellow seeds x r y R r Y y
R rYy
If two RrYy plants are crossed, the F2 generation would be:-
RY F1 genotypes: Gametes: RY Ry RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY Ry rY ry
RY F1 genotypes: Gametes: RY Ry RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY R Ry rY ry
RY F1 genotypes: Gametes: RY Ry RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RR Ry rY ry
RY F1 genotypes: Gametes: RY Ry RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RRY Ry rY ry
RY F1 genotypes: Gametes: RY Ry RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RRYY Ry rY ry
RY F1 genotypes: Gametes: RY Ry RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RRYY RRYy RrYY RrYy Ry RRYy RRyy RrYy Rryy rY RrYY RrYy rrYY rrYy ry RrYy Rryy rrYy rryy What is the phenotypic ratio?
RY F1 genotypes: Gametes: RY Ry RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RRYY RRYy RrYY RrYy Ry RRYy RRyy RrYy Rryy rY RrYY RrYy rrYY rrYy ry RrYy Rryy rrYy rryy R - round seed r - wrinkled seed Y - yellow seed y - green seed F2 phenotypes: round yellow › : round green £ : wrinkled yellow Õ : wrinkled green ¯ : R_Y_ R_yy rrY_ rryy
RY F1 genotypes: Gametes: RY Ry RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RRYY › RRYy › RrYY › RrYy › Ry RRYy › RRyy RrYy › Rryy rY RrYY › RrYy › rrYY rrYy ry RrYy › Rryy rrYy rryy F2 phenotypes: round yellow › : round green £ : wrinkled yellow Õ : wrinkled green ¯ : R_Y_ R_yy rrY_ rryy R - round seed r - wrinkled seed Y - yellow seed y - green seed
RY F1 genotypes: Gametes: RY Ry RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RRYY › RRYy › RrYY › RrYy › Ry RRYy › RRyy RrYy › Rryy rY RrYY › RrYy › rrYY rrYy ry RrYy › Rryy rrYy rryy F2 phenotypes: round yellow ›: round green £ : R_Y_ R_yy wrinkled yellow Õ :rrY_ wrinkled green ¯ : rryy R - round seed r - wrinkled seed Y - yellow seed y - green seed
RY F1 genotypes: Gametes: RY Ry RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RRYY › RRYy › RrYY › RrYy › Ry RRYy › RRyy £ RrYy › Rryy £ rY RrYY › RrYy › rrYY Õ rrYy Õ ry RrYy › Rryy £ rrYy Õ rryy ¯ F2 phenotypes: 9 round yellow ›: 3 round green £ : 3 wrinkled yellow Õ : 1 wrinkled green ¯ : R_Y_ R_yy rrY_ rryy
The phenotype ratio: 9 : 3 : 3 : 1 1) Parents are BOTH heterozygous for both genes. This ratio indicates: 2.The two genesare on separate chromosomes. x
What are the possible genotypes of this individuals gametes?
Bronze colour in turkey is controlled by a dominant allele, R, whereas red colour is controlled by are recessive allele, r. The dominant allele , H, produces normal feather while the recessive allele, h, produces ‘hairy` feather. In a cross between homozygous bronze ‘hairy`- feathered turkey and homozygous red , normal- feathered turkey, what are the fractions of F2 progeny with; a) Rrhh genotype? b) Bronze and ‘hairy`-feathered phenotype? c) rrHH genotype? d) Red and normal-feathered phenotype? e) RrHh genotype? f) Bronze and normal-feathered phenotype? g) RrHH genotype? h) RRHh ghenotype?
♀ ♂ RH Rh rH rh RH RRHH RRHh RrHH RrHh Rh RRHh RRhh RrHh Rrhh rH RrHH RrHh rrHH rrHh rh RrHh Rrhh rrHh rrhh
a) Rrhh genotype? 2/16 = ⅛ b) Bronze and ‘hairy`-feathered phonotype? 3/16 c) rrHH genotype? 1/16 d) Red and normal-feathered phenotype? 3/16 e) RrHh genotype? 4/16 = ¼ f) Bronze and normal-feathered phenotype? 9/16 g) RrHH genotype? 2/16 = ⅛ h) RRHh ghenotype? 2/16 = ⅛
▪ Let S – smooth (seed shape) s – wrinkled (seed shape) Y – yellow (seed colour) y – green (seed colour) SSYY SSYy SsYY SsYy Smooth seeds, yellow colour
sy SSYY SY x ssyy Parents: Gametes: SsYy x F1: (100% smooth & yellow) S – smooth (seed shape) s – wrinkled (seed shape) Y – yellow (seed colour) y – green (seed colour)
SY sy SSYY x ssyy Parents: Gametes: F1: SsYy (100% smooth & yellow) x
x ssyy Parents: Gametes: F1: sy SsYy SY Sy sY sy SsYy Ssyy ssYy ssyy 1/4 smooth : & yellow 1/4 smooth : 1/4 wrinkled : 1/4 wrinkled & green & yellow & green x S – smooth (seed shape) s – wrinkled (seed shape) Y – yellow (seed colour) y – green (seed colour)
SY SsYy x ssyy Parents: Gametes: F1: Sy sY sy sy Ssyy ssYy ssyy SsYy 1/4 smooth : & yellow 1/4 smooth : 1/4 wrinkled : 1/4 wrinkled & green & yellow & green x
x ssyy Parents: Gametes: F1: sy SSYy SY Sy SsYy Ssyy 1 smooth & yellow : 1 smooth & green x S – smooth (seed shape) s – wrinkled (seed shape) Y – yellow (seed colour) y – green (seed colour)
SY SSYy x ssyy Parents: Gametes: F1: Sy sy SsYy 1 smooth & yellow Ssyy : 1 smooth & green x
x ssyy Parents: Gametes F1: sy SsYY : SY sY SsYy ssYy 1 smooth & yellow : 1 wrinkled & yellow x S – smooth (seed shape) s – wrinkled (seed shape) Y – yellow (seed colour) y – green (seed colour)
SsYY x ssyy Parents: Gametes: F1: SY sY sy SsYy 1 smooth & yellow ssYy : 1 wrinkled & yellow x
i. Codominant Alleles ii. Incomplete Dominant Alleles iii. Multiple Alleles iv. Polygenes and Polygenic Inheritance v. Dominant Lethal Alleles vi. Linked Genes vii. Sex-Linked Genes
• When neither allele for a gene is recessive • Example: Blood type • AllelesA and B are both dominant (bothare expressed) • i is recessive to allelesA and B • One letter is chosen and the possible alleles are written in upper case letters to illustrate codominance Phenotypes Genotypes A IAIA orIAi B IBIB or IBi AB IAIB O ii
Neither allele is dominant. Heterozygous shows combined effects of both alleles i.e. intermediates/blending. Can produce other phenotype that is totally different from P.
Complete dominance Incomplete dominance The dominant allele completely masks the recessive one Neither allele is dominant RR rr Rr RR Rr rr
Red White Pink R allele: is partially dominant Neither allele is dominant. Heterozygous shows combined effects of both alleles i.e. intermediates/blending. Can produce other phenotype that is totally different from P.
F2 Phenotypic ratio: 1 white : 2 pink :1 red genotypic ratios are the same F2 Genotypicratio: 1 rr : 2 Rr :1 RR A ratio of 1:2:1 (in F2) is characteristic of INCOMPLETE DOMINANCE
▪ MultipleAlleles  Single trait may have more than 2 alleles ie multiple alleles/alternatives genes  Fruit fly is one example – eye colour  ABO blood group – 3 alleles IA, IB, i
 When the inheritance/expression of a characteristic is controlled by more than one gene.  Genes that responsible for continuous variation in human and animals ie height, skin colour.  Determined by a large number of genes at different loci.  Proteins produced by these genes will interact with each other to produce continuous variation in an organism.  Environment and diet will effect how genes being expressed.
▪ LethalAlleles  Allele that has negative affects on the survival of a homozygote i.e. causes death  Genes which result in the premature death of the organism; dominant lethal genes kill heterozygotes, whereas recessive lethal genes kill only homozygotes ▪ Two types: ▪ Dominant LethalAllele ▪ Autosomal dominant lethal disorder (both homo n hetero will be affected) ▪ Usually will not live long enough to reproduce ▪ Huntington’s disease,TaySach’sdisease ▪ Recessive LethalAllele ▪ Affect the survival of the organism usually as early as fetal stage ▪ Inviable (affect only homozygote)
▪ Epistasis occurs when the phenotypic expression of one gene is affected by another gene. ▪ In epistasis:  two genes interact to control a single phenotype ▪ not producing new phenotypes ▪ one modifies or masks the expression of the other ▪ The gene that masks another is epistatic ▪ The gene that is masked is hypostatic
Epistasis : Coat color in mice Yellow tip Black ▪ the epistatic gene: 1) controls synthesis of melanin 2) has two alleles:  coloured (M - dominant)  albino (m - recessive) ▪ the hypostatic gene: 1) controls distribution of pigment in the hair 2) its alleles are:  agouti (A - grey, dominant)  black (a - recessive) Black
MM or Mm mm AA orAa BLACK AGOUTI aa Colour forms Albino Epistatic gene (M): Melanin production Hypostatic gene (A): Melanin distribution
Epistasis : Coat color in mice Agouti Black Albino MMAA MMAa MmAA MmAa mmaa mmAA mmAa MMaa Mmaa
▪ Discovered by Morgan in the early 1900 when he crossed Drosophila. ▪ Do not conform to Mendel’s law of independent assortment. • Link genes are located on the same chromosome. • Genes inherited together during meiosis or as a single unit. • Do not sort independently.  But can the linked genes be separated?? ▪ Fail to produce the expected 9:3:3:1 ratio in a breeding situation involving dihybrid inheritance. ▪ In these situations a variety of ratios are produced.
▪ Genes for body colour and wing length are linked. Body colour:  grey [G]  black [g] Wing length:  long wings [L]  vestigial (short) wings [l] Vestigial wings G g L l
Parental genotypes: GGLL x ggll Gametes: GL x gl F1 genotypes: GgLl If F1 are allowed to interbreed: GgLl xGgLl Expected F2 phenotypes: 9:3:3:1 However, the F2 showed  3:1
3 grey body, long wing : 1 black body, vestigial wing
grey -G black - g long wings - L vestigial wings -l
▪ In practice:  this 3:1 ratio is never achieved reason: total linkage is rare [since crossing-over can occur] ▪ In reality:  4 phenotypes are produced
GgLl x ggll G – grey body g – black body L – long wing l – vestigial wing
Genotype GgLl Ggll ggLl ggll Phenotype grey, long wing grey, vestigial wing black, long wing black, vestigial wing Actual nos. 965 185 206 944 Expected 25% 25% 25% 25% Observed 41.5% 8.5% 8.5% 41.5%
Genotype GgLl Ggll ggLl ggll Phenotype grey, long wing grey, vestigial wing black, long wing black, vestigial wing Observed 41.5% 8.5% 8.5% 41.5% Like parents Unlike parents
Allow linked alleles to separate and recombine = forming new linkage ie recombinants
Let us explain how crossing-over may happen during gamete formation in:
G g L l Replication
G g L l Replication g L l G G g L l
G g L l Replication Crossing over g L l G G g L l
G g L l Replication Crossing over g L l G G g L l G L g l G g L l Recombinants
Grey body, long wing Greybody, vestigial wing Black body, vestigialwing Blackbody, long wing G – grey body g – black body L – long wing l – vestigial wing Recombinants
Grey body, long wing Greybody, vestigial wing Black body, vestigialwing Blackbody, long wing Recombinants 8.5% grey body, vestigial wing 8.5% black body, long wing Parental phenotypes Recombinants
1.approximately equal numbers of the parental phenotypes Grey Long wing Black vestigial Black Long wing Grey vestigial 965 944 206 185 2. a significantly smaller number of phenotypes showing new combinations of characteristics also in equal numbers, called recombinants
▪ COV – CrossOverValue  Shown by percentage of genes separation  No of recombinant offspring/Total no of offspring x 100%  1% = 1 map unit (mu) = 1 centimorgan (cM)  Can be used to determine the order in which genes are located on the chromosome
Recombinant frequency = Number of individuals showing recombination X 100 Number of offspring  parental phenotypes: grey body, long wing 965 black body, vestigial wing 944  recombinant phenotypes: black body, long wing grey body, vestigial wing 206 185 Recombinant frequency  100 17% 206 185 965  944 206 185
▪ Purpose:  Todetermine the linear order and distance of separation among genes that are linked to each other along the same chromosome ▪ Based to predict amount of crossing over  Greater the distance between two genes on a chromosome, the more likely crossing over occurs hence higher % of recombinants  Low percentage of recombinants will indicate that genes are relatively close together
▪ COV between genes into hypothetical distances along the chromosome C? A B C? 4 9 9 A COV of 4% : between genes A and B: they are 4 units apart A COV of 9% : for a pair of genes A andC: they are 9 units apart but does not indicate the linear sequence of genes
▪ consider the following values involving four genes, P,Q, R and S: P –Q = 24% R – P = 14% R – S = 8% S – P = 6%
▪ consider the following values involving four genes, P,Q, R and S: P –Q = 24% R – P = 14% R – S = 8% S – P = 6%
▪ consider the following values involving four genes, P,Q, R and S: P –Q = 24% R – P = 14% R – S = 8% S – P = 6%
▪ consider the following values involving four genes, P,Q, R and S: P – Q = 24% R – P = 14% R – S = 8% S – P = 6%
 Black body and vestigial wing in Drosophila is controlled by recessive alleles, b and vg, respectively.The dominant alleles, b+ and vg+, produce wild-type flies with grey body and normal wing. A homozygous wild type is crossed to a homozygous fly with black body and vestigial wing.TheF1 progeny (all wild-type) is test –crossed.The F2 generation is asfollows: ¡ ▪ 1930 wild type ▪ 1888 black, vestigial ▪ 412 black, normal ▪ 370 grey, vestigial ¡ a) Determine whether the genes controlling body colour and wing type are linked. b)If the genes are linked , determine the distance between the two genes. c) If the genes are not linked, what is the expected number of individuals of each phenotype in the F2 generation?
a) The genes are linked because the recombinant phenotype ratio is smaller compared to parental phenotype. b)No. of recombinants X 100 Total no. of progeny 412 + 370 x 100 = 17% = 17 map unit 4600 c) 1150
autosomes Sex chromosomes X Y Female carries two alleles of a gene X X Male carries one allele of a gene
 In case of defective genes ie colour blindness.  .  Why CB is only common in male and rarely in female? ▪ Father has defective genes onX. ▪ Male only receives X chromosome from his mother. ▪ Female receives X chromosome from her father. ▪ Hence the male won't have any defective genes but the female would become a carrier. ▪ Eventually, the defective genes would be transmitted to her sons.
•Examples of recessive sex-linked disorders: Colorblindness – inability to distinguish between certain colors most common type is red-green color blindness, where red and green are seen as the same color. Youshould see 58 (upper left), 18(upper right), E (lower left) and 17 (lower right).
hemophilia – blood won’t clot
Genotype Phenotype XHXH Normal female XHXh Normal female (carrier) XHY Normal male XhY Haemophiliac male
Pattern Baldness In Humans Baldness is an autosomal trait and is apparently influenced by sex hormones after people reach 30 years of age or older. In men the gene is dominant, while in women it is recessive.A man needs only one allele (B) for the baldness trait to be expressed, while a bald woman must be homozygous for the trait (BB). What are the probabilities for the children for a bald man and woman with no history of baldness in the family?
Normal A Females carry two allele of a gene. If one allele is defective, female is still normal as effect is masked by the normal allele. A Normal:A Sick: a Sick Phenotypically normal / carrier A A a a a a Normal Sick
▪ Is used to track inheritance patterns in families. ▪ What does a PEDIGREECHART show? The phenotypes of individuals in several generations of a family, and provides a basis for attempting to determine their genotype.
Pedigree analysis reveals Mendelian patterns in human inheritance •In these family trees, squares symbolize males and circles represent females. •A horizontal line connecting a male and female (--) indicates a mating, with offspring listed below in their order of birth, from left to right. •Shaded symbols stand for individuals with the trait being traced.
Key: AA = affected Aa = affected aa = normal = normal female = normal male = affected female = affected male Aa Aa aa Aa aa aa aa A? Aa aa Aa aa aa aa F1 generation F2 generation
1. A A
2. A A
3. Two affected parents can produce an unaffectedchild. A A
4. . A A
aa A? Aa A? Aa Aa Aa A? A? Key aa = affected Aa = carrier (appears normal) AA = normal A? aa A? aa Albinos
1. aa A? Aa A? Aa Aa Aa A? A? Key aa = affected Aa = carrier (appears normal) AA = normal A? aa A? aa
aa A? Aa A? Aa Aa Aa A? A? Key aa = affected Aa = carrier (appears normal) AA = normal A? aa A? aa 2.
3. Two affected parents will always haveaffected
4.
aa A? Aa A? Aa Aa Aa A? A? Key aa = affected Aa = carrier (appears normal) AA = normal A? aa A? aa 5.
1. Autosomal recessive  e.g. albinism 2. Autosomal dominant  e.g. Huntington’s Disease [degeneration of brain cells] 3. X-linked recessive  e.g. color-blindness, haemophilia 4. X-linked dominant [Very Rare]  e.g. hypophosphatemia [low level of phosphate in blood]
▪ Dominant traits- traits that are expressed. ▪ Recessive traits- traits that are covered up. ▪ Alleles- the different forms of a characteristic. ▪ Punnett Squares- show how crosses are made. ▪ Probability- the chances/ percentages that something will occur. ▪ Genotype- the types of genes (Alleles) present. ▪ Phenotype- what it looks like. ▪ Homozygous- two of the same alleles. ▪ Heterozygous- two different alleles.

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FBY 0416 - Chapter 4 - Genetic Inheritance (Latest).pptx

  • 1.
    CHAPTER 4 -GENETIC INHERITANCE
  • 3.
    At the endof the topics students should able to: 1. Define terminologies used in genetic inheritance. 2. State Mendel’s first law (law of segregation) and second law (law of independent assortment). 3. Calculate genotypic and phenotypic ratio. 4. Explain codominant alleles, incomplete dominant alleles, multiple alleles, linked genes and sex-linked genes. 5. Calculate genetic distance (map unit).
  • 4.
    Genetic inheritance Terms & concepts Mendel’s experiment Pisumsativum – garden pea Monohybrid crosses Dihybrid crosses Law of segregation Law of independent assortment Deviation from Mendel’s Law Codominance Incomplete dominance Multiple allele Polygene Lethal allele Linked gene Genetic mapping Pedigree analysis
  • 5.
    What is agene? A. A factor that controls a heritable characteristic B. Something on a chromosome C. Information stored in a segment of DNA D. Something that encodes a protein
  • 6.
    ▪ Inheritance ishow traits, or characteristics, are passed on from generation to generation. ▪ Chromosomes are made up of genes, which are made up of DNA. ▪ Genetic material (genes, chromosomes, DNA) is found inside the nucleus of acell. ▪ Particulate Hypothesis of Inheritance. ▪ Parents pass on to their offspring separate and distinct factors (today called genes) that are responsible for inherited traits.
  • 7.
    ▪ Gene -A piece of DNA that encodes a particular trait. EX a gene for eye-color ▪ Allele-an alternate form of a gene.  E.g. - allele for blue eyes and allele for brown eyes ▪ Locus — the location of a gene on a chromosome. Plural =loci ▪ Hybrid -The offspring of genetically dissimilarfrom parents. ▪ Phenotype-the physical appearance of an individual, determined by his or her genotype. (black, brown, short, tall, etc). ▪ Genotype-the genetic composition of an individual or combination of alleles. (BB, Bb, bb)
  • 8.
     Homozygous -containing a pair of the same alleles.Can be ▪ Homozygous recessive - two recessive alleles e.g. bb, or ▪ Homozygous dominant – two dominant alleles e.g. BB ▪ Heterozygous - containing two different alleles. E.g. Bb ▪ Carrier – an individual who has a recessive allele of a gene that does not have an effect on their phenotype. ▪ Dominant allele — expressed whether alone or in pairs. Symbolized by a capital letter.  E.g. Brown eye allele=B ▪ Recessive allele — expressed only in the absence of a dominant allele. Symbolized by lowercase.  E.g. blue eye allele=b
  • 10.
    ▪ Austrian Monk. ▪Is considered “The Father of Genetics" ▪ Experimented with “pea plants”. ▪ Used pea plants because they:  were available  reproduced quickly  showed obvious differences in the traits Understood that there was something that carried traits from one generation to the next- “FACTOR”.
  • 12.
    1. Law ofDominance: - In a cross of parents that are pure for contrasting traits, only one form of the trait will appear in the next generation. - Offspring that are hybrid for a trait will have only the dominant trait in the phenotype. What happens when the F1’s are crossed?
  • 13.
    The F1 crossedproduced the F2 generation and the lost trait appeared with predictable ratios. This led to the formulation of the current model of inheritance. ● In a hybrid union, the allele which expresses itself phenotypically = dominant allele while the other allele which fails to express itself phenotypically = recessive allele. ● The hybrid individual shows phenotypically only the dominant character. ● The law of dominance is often described as Mendel’s first law of inheritance.
  • 15.
    2. Law ofSegregations: -During the formation of gametes (eggs or sperm), the two alleles (hereditary units) responsible for a trait separate from each other. -Alleles for a trait are then "recombined" at fertilization, producing the genotype for the traits of the offspring. 3. Law of IndependentAssortment: - Alleles for different traits are distributed to sex cells(& offspring) independently of one another.
  • 16.
    ▪A hybrid producedby crossing t wo individuals with different alleles at one genetic locus of interest. ▪In Mendel's experiments with garden peas (Pisum sativum) . ▪ Earliest and simplest breeding experiment. ▪ Organisms showed contrasting traits in character were crossed.
  • 18.
    1. they wereeasy to grow 2. they had a short life cycle 3. their pollination could be controlled 4. they have easily observable characteristics.
  • 19.
    He studied 7characteristics, each of which has two contrasting alternatives. seed shape : round / wrinkled pod color : yellow / green seed color : yellow / green flower color : purple / white pod shape : inflated / constricted flower position : axial / terminal plant height : tall / dwarf
  • 20.
    Totest the particulatehypothesis, Mendel crossed true-breeding plants that had two distinct and contrasting and found that the resulting offspring ALWAYSlooked like just one of the parents, not a combination of the two. What is meant by “true breeding?”
  • 21.
    True breed organism–homozygous dominant or homozygous recessive. ▪ A true breeding organism or a purebred, is an organism that always •2 true-breeding purple flower(PP) is crossed with white flower(pp). •When self-pollinate or crossed among themselves – all offspring produced will be identical to the parent which all have purple flower.
  • 22.
    ▪ This 1stcrossing is calledhybridisation ▪ True breeding parents are called – P generation ▪ Their offspring are called – F1 generation or (First Filial Generation) ▪ Next he allowed the F1 generation to self-pollinate ▪ This will produce the Second Filial Generation orF2 generation. What happened ??
  • 23.
    Conclusion: Recessive traithad not been destroyed or deleted in F1 generation, but merely masked by dominant trait.
  • 24.
    ▪ 4 concepts: 1.Alternative versions of genes account for variations in inherited characters.  Gene for flower colour exists in 2 versions: ▪ Purple and white i.e. P and p. ▪ These alternative versions of a gene are called Alleles. ▪ Location of a gene on a chromosome is called Locus.
  • 25.
    Alleles: alternative versionsof a gene. The gene for a particular inherited character resides at a specific locus (position) on homologous chromosome. For each character, an organism inherits two alleles, one from each parent.
  • 26.
    2.  The 2alleles at a particular locus may be identical (true-breeding plants,TT or tt) or it may differ as in the F1 hybrids(Tt) 3. If the 2 alleles at a locus differ, then one, the dominant allele determines the organism’s appearance, the other, the recessive allele, has no noticeable effect on the organism. 4. The 2 alleles for a heritable character segregate (separate) during gamete formation and end up in different gametes. This is the Law of Segregation orThe 1st Mendel’sLaw.
  • 27.
    ▪ Two allelesfor a gene segregate during gamete formation and are rejoined at random, one from each parent, during fertilization.
  • 28.
    ▪ Used todetermine the outcome of a cross between two individuals.
  • 29.
    How does agenotype ratio differ from the phenotype ratio?
  • 30.
    When any individualproduces gametes, the alleles separate, so that each gamete receives only one member of the pair of alleles. ▪ An exception to this rule is linked genes Gametes:
  • 32.
    Is designed toreveal whether an organism that displays the dominant phenotype is homozygous or heterozygous. A cross between a recessive homozygous and an organism of dominant phenotype, but unknown genotype.  If an organism displays a dominant characteristic, it may possess 2 dominant alleles (homologous) or a dominant and recessive allele for that characteristic (heterozygous)  To find out which in the case, the organism is crossed with one displaying the recessive characteristic.  If all the offspring show the dominant characteristic then the organism is homozygous, but if half show the recessive characteristic, then the organism is heterozygous.
  • 34.
     A matingbetween individuals of the parental generation (P) and the first generation (F1). ▪ A backcross can be made to the dominant parental type or to the recessive parental type. ▪ If the backcross is to the dominant parent, all offspring show the dominant phenotype. ▪ If the backcross is to the recessive parent (a testcross), ½ the offspring have the dominant phenotype and ½ have the recessive phenotype.
  • 35.
    ▪ A backcross=test cross if a parent is homozygous recessive.
  • 36.
     A crossreversing the roles of males and females to confirm the results obtained from an earlier cross. @  A cross with the phenotype of each sex reversed as compared with the original cross.
  • 37.
    ▪ Each pairof alleles segregates independently during gamete formation.  Allele for one gene will be found within a resulting gamete independently of the allele for a different gene in the same gamete. ▪ Mendel experimented this using 2 pairs of contrasting traits at a time i.e. dihybrid cross. ▪ This law applies only to genes (allele pairs) located on different chromosomes. ▪ Genes located near each other or in the same chromosome tend to be inherited together and have more complex inheritance patterns.
  • 38.
    ▪ Each pairof alleles segregates independently during gamete formation.
  • 39.
    It applies togenes that lie on separate chromosomes. It does not apply when genes lie on the same chromosome.
  • 40.
    ▪ The inheritanceof two separate traits in a single cross ▪ for example: RRYY x rryy Let: R represent round seed r represent wrinkled seed Y represent yellow seed y represent green seed Seed shape Seed colour
  • 41.
    Parental Phenotypes: Wrinkled, seeds Parental genotypes:rryy Round, yellow seeds RRYY x x green R R Y Y r r y y Gametes: R Y F1 genotypes: F1 phenotypes: 100% round, yellow seeds x r y R r Y y
  • 42.
  • 43.
    If two RrYyplants are crossed, the F2 generation would be:-
  • 44.
    RY F1 genotypes: Gametes: RYRy RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY Ry rY ry
  • 45.
    RY F1 genotypes: Gametes: RYRy RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY R Ry rY ry
  • 46.
    RY F1 genotypes: Gametes: RYRy RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RR Ry rY ry
  • 47.
    RY F1 genotypes: Gametes: RYRy RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RRY Ry rY ry
  • 48.
    RY F1 genotypes: Gametes: RYRy RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RRYY Ry rY ry
  • 49.
    RY F1 genotypes: Gametes: RYRy RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RRYY RRYy RrYY RrYy Ry RRYy RRyy RrYy Rryy rY RrYY RrYy rrYY rrYy ry RrYy Rryy rrYy rryy What is the phenotypic ratio?
  • 50.
    RY F1 genotypes: Gametes: RYRy RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RRYY RRYy RrYY RrYy Ry RRYy RRyy RrYy Rryy rY RrYY RrYy rrYY rrYy ry RrYy Rryy rrYy rryy R - round seed r - wrinkled seed Y - yellow seed y - green seed F2 phenotypes: round yellow › : round green £ : wrinkled yellow Õ : wrinkled green ¯ : R_Y_ R_yy rrY_ rryy
  • 51.
    RY F1 genotypes: Gametes: RYRy RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RRYY › RRYy › RrYY › RrYy › Ry RRYy › RRyy RrYy › Rryy rY RrYY › RrYy › rrYY rrYy ry RrYy › Rryy rrYy rryy F2 phenotypes: round yellow › : round green £ : wrinkled yellow Õ : wrinkled green ¯ : R_Y_ R_yy rrY_ rryy R - round seed r - wrinkled seed Y - yellow seed y - green seed
  • 52.
    RY F1 genotypes: Gametes: RYRy RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RRYY › RRYy › RrYY › RrYy › Ry RRYy › RRyy RrYy › Rryy rY RrYY › RrYy › rrYY rrYy ry RrYy › Rryy rrYy rryy F2 phenotypes: round yellow ›: round green £ : R_Y_ R_yy wrinkled yellow Õ :rrY_ wrinkled green ¯ : rryy R - round seed r - wrinkled seed Y - yellow seed y - green seed
  • 53.
    RY F1 genotypes: Gametes: RYRy RrYy rY ry RrYy x Ry rY ry x ♂ ♀ RY Ry rY ry RY RRYY › RRYy › RrYY › RrYy › Ry RRYy › RRyy £ RrYy › Rryy £ rY RrYY › RrYy › rrYY Õ rrYy Õ ry RrYy › Rryy £ rrYy Õ rryy ¯ F2 phenotypes: 9 round yellow ›: 3 round green £ : 3 wrinkled yellow Õ : 1 wrinkled green ¯ : R_Y_ R_yy rrY_ rryy
  • 54.
    The phenotype ratio:9 : 3 : 3 : 1 1) Parents are BOTH heterozygous for both genes. This ratio indicates: 2.The two genesare on separate chromosomes. x
  • 57.
    What are thepossible genotypes of this individuals gametes?
  • 59.
    Bronze colour inturkey is controlled by a dominant allele, R, whereas red colour is controlled by are recessive allele, r. The dominant allele , H, produces normal feather while the recessive allele, h, produces ‘hairy` feather. In a cross between homozygous bronze ‘hairy`- feathered turkey and homozygous red , normal- feathered turkey, what are the fractions of F2 progeny with; a) Rrhh genotype? b) Bronze and ‘hairy`-feathered phenotype? c) rrHH genotype? d) Red and normal-feathered phenotype? e) RrHh genotype? f) Bronze and normal-feathered phenotype? g) RrHH genotype? h) RRHh ghenotype?
  • 60.
    ♀ ♂ RH Rh rHrh RH RRHH RRHh RrHH RrHh Rh RRHh RRhh RrHh Rrhh rH RrHH RrHh rrHH rrHh rh RrHh Rrhh rrHh rrhh
  • 61.
    a) Rrhh genotype?2/16 = ⅛ b) Bronze and ‘hairy`-feathered phonotype? 3/16 c) rrHH genotype? 1/16 d) Red and normal-feathered phenotype? 3/16 e) RrHh genotype? 4/16 = ¼ f) Bronze and normal-feathered phenotype? 9/16 g) RrHH genotype? 2/16 = ⅛ h) RRHh ghenotype? 2/16 = ⅛
  • 63.
    ▪ Let S –smooth (seed shape) s – wrinkled (seed shape) Y – yellow (seed colour) y – green (seed colour) SSYY SSYy SsYY SsYy Smooth seeds, yellow colour
  • 64.
    sy SSYY SY x ssyy Parents: Gametes: SsYy x F1: (100% smooth& yellow) S – smooth (seed shape) s – wrinkled (seed shape) Y – yellow (seed colour) y – green (seed colour)
  • 65.
    SY sy SSYY xssyy Parents: Gametes: F1: SsYy (100% smooth & yellow) x
  • 66.
    x ssyy Parents: Gametes: F1: sy SsYy SY SysY sy SsYy Ssyy ssYy ssyy 1/4 smooth : & yellow 1/4 smooth : 1/4 wrinkled : 1/4 wrinkled & green & yellow & green x S – smooth (seed shape) s – wrinkled (seed shape) Y – yellow (seed colour) y – green (seed colour)
  • 67.
    SY SsYy x ssyy Parents: Gametes: F1: SysY sy sy Ssyy ssYy ssyy SsYy 1/4 smooth : & yellow 1/4 smooth : 1/4 wrinkled : 1/4 wrinkled & green & yellow & green x
  • 68.
    x ssyy Parents: Gametes: F1: sy SSYy SY Sy SsYySsyy 1 smooth & yellow : 1 smooth & green x S – smooth (seed shape) s – wrinkled (seed shape) Y – yellow (seed colour) y – green (seed colour)
  • 69.
    SY SSYy x ssyy Parents: Gametes: F1: Sysy SsYy 1 smooth & yellow Ssyy : 1 smooth & green x
  • 70.
    x ssyy Parents: Gametes F1: sy SsYY : SY sY SsYyssYy 1 smooth & yellow : 1 wrinkled & yellow x S – smooth (seed shape) s – wrinkled (seed shape) Y – yellow (seed colour) y – green (seed colour)
  • 71.
    SsYY x ssyy Parents: Gametes: F1: SYsY sy SsYy 1 smooth & yellow ssYy : 1 wrinkled & yellow x
  • 72.
    i. Codominant Alleles ii.Incomplete Dominant Alleles iii. Multiple Alleles iv. Polygenes and Polygenic Inheritance v. Dominant Lethal Alleles vi. Linked Genes vii. Sex-Linked Genes
  • 73.
    • When neitherallele for a gene is recessive • Example: Blood type • AllelesA and B are both dominant (bothare expressed) • i is recessive to allelesA and B • One letter is chosen and the possible alleles are written in upper case letters to illustrate codominance Phenotypes Genotypes A IAIA orIAi B IBIB or IBi AB IAIB O ii
  • 74.
    Neither allele isdominant. Heterozygous shows combined effects of both alleles i.e. intermediates/blending. Can produce other phenotype that is totally different from P.
  • 75.
    Complete dominance Incompletedominance The dominant allele completely masks the recessive one Neither allele is dominant RR rr Rr RR Rr rr
  • 76.
    Red White Pink R allele: ispartially dominant Neither allele is dominant. Heterozygous shows combined effects of both alleles i.e. intermediates/blending. Can produce other phenotype that is totally different from P.
  • 77.
    F2 Phenotypic ratio: 1white : 2 pink :1 red genotypic ratios are the same F2 Genotypicratio: 1 rr : 2 Rr :1 RR A ratio of 1:2:1 (in F2) is characteristic of INCOMPLETE DOMINANCE
  • 78.
    ▪ MultipleAlleles  Singletrait may have more than 2 alleles ie multiple alleles/alternatives genes  Fruit fly is one example – eye colour  ABO blood group – 3 alleles IA, IB, i
  • 79.
     When theinheritance/expression of a characteristic is controlled by more than one gene.  Genes that responsible for continuous variation in human and animals ie height, skin colour.  Determined by a large number of genes at different loci.  Proteins produced by these genes will interact with each other to produce continuous variation in an organism.  Environment and diet will effect how genes being expressed.
  • 81.
    ▪ LethalAlleles  Allelethat has negative affects on the survival of a homozygote i.e. causes death  Genes which result in the premature death of the organism; dominant lethal genes kill heterozygotes, whereas recessive lethal genes kill only homozygotes ▪ Two types: ▪ Dominant LethalAllele ▪ Autosomal dominant lethal disorder (both homo n hetero will be affected) ▪ Usually will not live long enough to reproduce ▪ Huntington’s disease,TaySach’sdisease ▪ Recessive LethalAllele ▪ Affect the survival of the organism usually as early as fetal stage ▪ Inviable (affect only homozygote)
  • 83.
    ▪ Epistasis occurswhen the phenotypic expression of one gene is affected by another gene. ▪ In epistasis:  two genes interact to control a single phenotype ▪ not producing new phenotypes ▪ one modifies or masks the expression of the other ▪ The gene that masks another is epistatic ▪ The gene that is masked is hypostatic
  • 84.
    Epistasis : Coatcolor in mice Yellow tip Black ▪ the epistatic gene: 1) controls synthesis of melanin 2) has two alleles:  coloured (M - dominant)  albino (m - recessive) ▪ the hypostatic gene: 1) controls distribution of pigment in the hair 2) its alleles are:  agouti (A - grey, dominant)  black (a - recessive) Black
  • 85.
    MM or Mm mm AAorAa BLACK AGOUTI aa Colour forms Albino Epistatic gene (M): Melanin production Hypostatic gene (A): Melanin distribution
  • 86.
    Epistasis : Coatcolor in mice Agouti Black Albino MMAA MMAa MmAA MmAa mmaa mmAA mmAa MMaa Mmaa
  • 88.
    ▪ Discovered byMorgan in the early 1900 when he crossed Drosophila. ▪ Do not conform to Mendel’s law of independent assortment. • Link genes are located on the same chromosome. • Genes inherited together during meiosis or as a single unit. • Do not sort independently.  But can the linked genes be separated?? ▪ Fail to produce the expected 9:3:3:1 ratio in a breeding situation involving dihybrid inheritance. ▪ In these situations a variety of ratios are produced.
  • 89.
    ▪ Genes forbody colour and wing length are linked. Body colour:  grey [G]  black [g] Wing length:  long wings [L]  vestigial (short) wings [l] Vestigial wings G g L l
  • 90.
    Parental genotypes: GGLLx ggll Gametes: GL x gl F1 genotypes: GgLl If F1 are allowed to interbreed: GgLl xGgLl Expected F2 phenotypes: 9:3:3:1 However, the F2 showed  3:1
  • 91.
    3 grey body, longwing : 1 black body, vestigial wing
  • 93.
    grey -G black -g long wings - L vestigial wings -l
  • 94.
    ▪ In practice: this 3:1 ratio is never achieved reason: total linkage is rare [since crossing-over can occur] ▪ In reality:  4 phenotypes are produced
  • 95.
    GgLl x ggll G– grey body g – black body L – long wing l – vestigial wing
  • 98.
    Genotype GgLl GgllggLl ggll Phenotype grey, long wing grey, vestigial wing black, long wing black, vestigial wing Actual nos. 965 185 206 944 Expected 25% 25% 25% 25% Observed 41.5% 8.5% 8.5% 41.5%
  • 99.
    Genotype GgLl GgllggLl ggll Phenotype grey, long wing grey, vestigial wing black, long wing black, vestigial wing Observed 41.5% 8.5% 8.5% 41.5% Like parents Unlike parents
  • 100.
    Allow linked allelesto separate and recombine = forming new linkage ie recombinants
  • 101.
    Let us explainhow crossing-over may happen during gamete formation in:
  • 102.
  • 103.
  • 104.
    G g L l Replication Crossingover g L l G G g L l
  • 105.
    G g L l Replication Crossingover g L l G G g L l G L g l G g L l Recombinants
  • 106.
    Grey body, long wing Greybody, vestigial wing Blackbody, vestigialwing Blackbody, long wing G – grey body g – black body L – long wing l – vestigial wing Recombinants
  • 107.
    Grey body, long wing Greybody, vestigial wing Blackbody, vestigialwing Blackbody, long wing Recombinants 8.5% grey body, vestigial wing 8.5% black body, long wing Parental phenotypes Recombinants
  • 108.
    1.approximately equal numbers ofthe parental phenotypes Grey Long wing Black vestigial Black Long wing Grey vestigial 965 944 206 185 2. a significantly smaller number of phenotypes showing new combinations of characteristics also in equal numbers, called recombinants
  • 109.
    ▪ COV –CrossOverValue  Shown by percentage of genes separation  No of recombinant offspring/Total no of offspring x 100%  1% = 1 map unit (mu) = 1 centimorgan (cM)  Can be used to determine the order in which genes are located on the chromosome
  • 110.
    Recombinant frequency = Numberof individuals showing recombination X 100 Number of offspring  parental phenotypes: grey body, long wing 965 black body, vestigial wing 944  recombinant phenotypes: black body, long wing grey body, vestigial wing 206 185 Recombinant frequency  100 17% 206 185 965  944 206 185
  • 111.
    ▪ Purpose:  Todeterminethe linear order and distance of separation among genes that are linked to each other along the same chromosome ▪ Based to predict amount of crossing over  Greater the distance between two genes on a chromosome, the more likely crossing over occurs hence higher % of recombinants  Low percentage of recombinants will indicate that genes are relatively close together
  • 112.
    ▪ COV betweengenes into hypothetical distances along the chromosome C? A B C? 4 9 9 A COV of 4% : between genes A and B: they are 4 units apart A COV of 9% : for a pair of genes A andC: they are 9 units apart but does not indicate the linear sequence of genes
  • 113.
    ▪ consider thefollowing values involving four genes, P,Q, R and S: P –Q = 24% R – P = 14% R – S = 8% S – P = 6%
  • 114.
    ▪ consider thefollowing values involving four genes, P,Q, R and S: P –Q = 24% R – P = 14% R – S = 8% S – P = 6%
  • 115.
    ▪ consider thefollowing values involving four genes, P,Q, R and S: P –Q = 24% R – P = 14% R – S = 8% S – P = 6%
  • 116.
    ▪ consider thefollowing values involving four genes, P,Q, R and S: P – Q = 24% R – P = 14% R – S = 8% S – P = 6%
  • 118.
     Black bodyand vestigial wing in Drosophila is controlled by recessive alleles, b and vg, respectively.The dominant alleles, b+ and vg+, produce wild-type flies with grey body and normal wing. A homozygous wild type is crossed to a homozygous fly with black body and vestigial wing.TheF1 progeny (all wild-type) is test –crossed.The F2 generation is asfollows: ¡ ▪ 1930 wild type ▪ 1888 black, vestigial ▪ 412 black, normal ▪ 370 grey, vestigial ¡ a) Determine whether the genes controlling body colour and wing type are linked. b)If the genes are linked , determine the distance between the two genes. c) If the genes are not linked, what is the expected number of individuals of each phenotype in the F2 generation?
  • 119.
    a) The genesare linked because the recombinant phenotype ratio is smaller compared to parental phenotype. b)No. of recombinants X 100 Total no. of progeny 412 + 370 x 100 = 17% = 17 map unit 4600 c) 1150
  • 120.
    autosomes Sex chromosomes X Y Female carriestwo alleles of a gene X X Male carries one allele of a gene
  • 122.
     In caseof defective genes ie colour blindness.  .  Why CB is only common in male and rarely in female? ▪ Father has defective genes onX. ▪ Male only receives X chromosome from his mother. ▪ Female receives X chromosome from her father. ▪ Hence the male won't have any defective genes but the female would become a carrier. ▪ Eventually, the defective genes would be transmitted to her sons.
  • 123.
    •Examples of recessivesex-linked disorders: Colorblindness – inability to distinguish between certain colors most common type is red-green color blindness, where red and green are seen as the same color. Youshould see 58 (upper left), 18(upper right), E (lower left) and 17 (lower right).
  • 124.
  • 125.
    Genotype Phenotype XHXH Normalfemale XHXh Normal female (carrier) XHY Normal male XhY Haemophiliac male
  • 126.
    Pattern Baldness InHumans Baldness is an autosomal trait and is apparently influenced by sex hormones after people reach 30 years of age or older. In men the gene is dominant, while in women it is recessive.A man needs only one allele (B) for the baldness trait to be expressed, while a bald woman must be homozygous for the trait (BB). What are the probabilities for the children for a bald man and woman with no history of baldness in the family?
  • 127.
    Normal A Females carry two alleleof a gene. If one allele is defective, female is still normal as effect is masked by the normal allele. A Normal:A Sick: a Sick Phenotypically normal / carrier A A a a a a Normal Sick
  • 128.
    ▪ Is usedto track inheritance patterns in families. ▪ What does a PEDIGREECHART show? The phenotypes of individuals in several generations of a family, and provides a basis for attempting to determine their genotype.
  • 129.
    Pedigree analysis revealsMendelian patterns in human inheritance •In these family trees, squares symbolize males and circles represent females. •A horizontal line connecting a male and female (--) indicates a mating, with offspring listed below in their order of birth, from left to right. •Shaded symbols stand for individuals with the trait being traced.
  • 130.
    Key: AA = affected Aa= affected aa = normal = normal female = normal male = affected female = affected male Aa Aa aa Aa aa aa aa A? Aa aa Aa aa aa aa F1 generation F2 generation
  • 131.
  • 132.
  • 133.
    3. Two affectedparents can produce an unaffectedchild. A A
  • 134.
  • 135.
    aa A? Aa A? Aa AaAa A? A? Key aa = affected Aa = carrier (appears normal) AA = normal A? aa A? aa Albinos
  • 136.
    1. aa A? Aa A? Aa AaAa A? A? Key aa = affected Aa = carrier (appears normal) AA = normal A? aa A? aa
  • 137.
    aa A? Aa A? Aa AaAa A? A? Key aa = affected Aa = carrier (appears normal) AA = normal A? aa A? aa 2.
  • 138.
    3. Two affectedparents will always haveaffected
  • 139.
  • 140.
    aa A? Aa A? Aa AaAa A? A? Key aa = affected Aa = carrier (appears normal) AA = normal A? aa A? aa 5.
  • 141.
    1. Autosomal recessive e.g. albinism 2. Autosomal dominant  e.g. Huntington’s Disease [degeneration of brain cells] 3. X-linked recessive  e.g. color-blindness, haemophilia 4. X-linked dominant [Very Rare]  e.g. hypophosphatemia [low level of phosphate in blood]
  • 142.
    ▪ Dominant traits-traits that are expressed. ▪ Recessive traits- traits that are covered up. ▪ Alleles- the different forms of a characteristic. ▪ Punnett Squares- show how crosses are made. ▪ Probability- the chances/ percentages that something will occur. ▪ Genotype- the types of genes (Alleles) present. ▪ Phenotype- what it looks like. ▪ Homozygous- two of the same alleles. ▪ Heterozygous- two different alleles.