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Genetics Two: Mendel Genetics Two: Mendel Presentation Transcript

  • Topic 9: Mendelian Genetics
  • Principle of Segregation: Each Parent or Gamete Contributes One Allele to Offspring Gregory Mendel's research into the breeding of pea plants was the first illustration of his genetic laws.
  • Mendel’s 1st Law Law of Segregation
    • Four parts
    • Alternative versions of genes account for variations in inherited characteristics.
    • For each characteristic, an organism inherits two alleles, one from each parent.
    • If the two alleles differ, then one, the allele that encodes the dominant trait, is fully expressed in the organism's appearance; the other, the allele encoding the recessive trait, has no noticeable effect on the organism's appearance.
    • The two alleles for each characteristic segregate during gamete production
    • Genotype is the combination of alleles for a gene that an individual.
    • Phenotype is smooth coat seed is crossed with rough coated seed
    • . In the smooth coated individual the alleles are identical this is called HOMOZYGOUS . Both parents are homozygous.
    • Alleles are segregated by meiosis. For this allele all gametes from one individual are the same.
    •  
    • Fertilization brings together the allele from one parent with the allele from the other parent to form a new homologous pair
    • F1 genotype or first filial generation have different allele this is called
    • HETEROZYGOUS in the example to the left.
    • In the example, the F1 Phenotype have a smooth coated seeds.
    • Punnet Square :
    • Used to determine the outcome of a cross between two individuals. In the example we have two parents that are heterozygous dominant for a trait
    • P p
    • P PP Pp
    • p Pp pp
    Offspring: Genotype : 1/4 PP, 1/2 Pp, and 1/4 pp Phenotype : 3/4 Purple and 1/4 white
  •  
  • Mendel's Second Law Law of Independent Assortment
    • This law states that allele pairs separate independently during the formation of gametes.
    • Therefore, traits are transmitted to offspring independently of one another
  •  
  • Mendel's laws of Independent Assortment and Meiosis
    • Meiosis creates genetic variation in the gametes of
    • an individual. Combinations of genes occur in the
    • gametes that do not occur in the parental organism
    • This is do to crossing over that occurs during...
    • Prophase I:
    • New genetic variation are created on linked genes (The matching chromosomes or pairs of homologous
    • chromosomes)
    • Metaphase II:
    • Genetic variation of unlinked genes (different chromosomes) by random assortment of the homologous pairs.
  • Recombinants from Cross over
    • The diagram below shows the loci of genes A and B.
    • The genes are on the same chromosome (A and B are a linkage group)
  • Recombinants from Cross over continued
    • The cross over involves the exchange of lengths of DNA.
    • In doing so the chromosomes of a homologous pair exchange their alleles.
  • Recombinants from Cross over continued
    • When the chromatids separate at anaphase II this will produce new combinations of alleles in the gametes (recombinants).
    • The recombinants in this example are aB and Ab
  • 8.3.3 Linkage group:
    • genes on the same chromosome
    • genes inherited together
    • genes that do not show the expected Mendelian ratios as predicted by the Laws Independent Assortment.
    The genes A and B are a linkage group. If this genotype (AaBb) was crossed with itself would produce a 3:1 ratio not a 9:3:3:1
  • Recombination
    • Recombination occurs with the re assortment of genes
    • or characters into different combinations from those of
    • the parents.
    • Causes of Recombination
    • 1. Unlinked Genes ( Prophase I)
    • Unlinked genes are not on the same chromosomes.
    • Recombination occurs by chromosome assortment.
  • Recombination
    • 2. Linked Genes (Metaphase II)
    • Linked genes are on the same chromosome.
    • They tend to be inherited together.
    • New combinations are caused by crossover
  • Autosomes and Sex Chromosomes
    • Autosomes
    • Autosomes are all the chromosomes but not the X and Y, the so called sex chromosomes
    • Sex Chromosomes
    • XY chromosomes, XX in female and XY in males. gene on the chromosomes may be sex linked
  • Example of Mendel’s Second Law
    • Gender determination
      • Sex Chromosomes determine the gender of an individual
      • Females have two X chromosomes.
      • The X-chromosome is fairly large ands carries many genes.
      • Males have one X chromosome and one Y chromosome.
      • The Y chromosome is shorter than the X chromosome and carried fewer genes.
      • Females inherit one X from their mother and one X from their father.
      • Male inherit their X chromosome from their mother and the Y from their father.
  • Inheritance of gender:
    • The ovum (egg) always carries the X chromosome.
    • The sperm either carries the X chromosome or the Y chromosome.
    • It is therefore the male gamete that determines the gender.
    • The probability of the egg carrying the X chromosome is 1.0
    • The probability of the male carrying the X chromosome is 0.5
    • Therefore the chance of and XX fertilization is 1.0 x 0.5 = 0.5
  • Homologous and Non-Homologous regions of the Sex Chromosomes
    • For the male sex chromosomes their are
    • non-homologous region males in which there is only one allele per gene and that is inherited from the female on the X-chromosome
    • In the homologous region the male inherited two copies of an allele per gene. As per the normal situation.
  • Homologous and Non-Homologous regions of the Sex Chromosomes
    • On the female sex chromosomes all regions of the X chromosome are homologous.
    • There are two alleles per gene as with all other genes on all other chromosomes
    This difference in x and y chromosomes plays a large role in determining rates of genetic inherited defects
  • Sex Linkage Alleles on the non-homologous region of the X chromosome are more common in females than in males
    • A gene with two alleles where one is dominant and one is recessive.
    • Female has three possible genotypes and one is the homozygous recessive.
    • In a population the chance of being homozygous recessive is 33.3 %.
    • Males have two possible genotypes.
    • There is a 50% chance of the homozygous recessive condition in the population.
    • In sex linked conditions the recessive condition is more common in males than females.
  • Sex Linkage Alleles on the non-homologous region of the X chromosome are more common in females than in males
  • Sex Linkage Examples
    • Hemophilia is an example of a sex linkage condition.
    • The hemophilia allele is recessive to the normal allele.
    • The gene is located on the non-homologous region of the X chromosome.
    • The disease is associated with an inability to produce a clotting factor in blood.
    • Internal bleeding takes longer to stop.
  • Sex Linkage Examples
    • The homozygous genotype(*) in females has a high mortality.
    • The genotype X n Y in males has a high mortality.
  • Sex Linkage Examples
    • Red Green Color Blindness is an example of a sex linked condition.
    • Red Green Color blindness is a recessive condition.
    • The color blind allele is recessive to the normal allele.
    • Female homozygous recessives X b X b are color blind.
    • Males with the genotype X b Y are color blind.
    • Notice that in a population the probability of having a Red Green color blind genotype in males is higher.
  • Sex Linkage Examples
  • Females (homogametic sex) and X-linked alleles
    • Human females are homogametic that is their 23 rd pair of
    • chromosomes are identical and called the XX chromosomes.
    • Human females can be heterozygous for sex linked alleles.
    • e.g.
          • Hemophilia X H X h
            • or
            • Color Blindness X B X b
    • Human females can be homozygous for sex linked alleles
    • e.g.
    • Hemophilia X h X h
    • or
    • Color Blindness X b X b
  • Female carriers of sex linked alleles
    • Female heterozygote's for sex linked alleles e.g. Hemophilia X H X h or Color Blindness X B X b are carriers of the allele.
    • They are unaffected by the condition.
    • They do pass on the allele which may result in a homozygous female or a male with the sex linked recessive allele.
  • Monohybrid crosses
    • Allele Keys: For dominant use upper-case (A) and for recessive alleles try to and lower-case letters(a) that are distinctive.Try to calculate and predict the genotypic and phenotypic ratio of offspring from the following crosses.
  • Example of a non sex linked cross: Cystic fibrosis (CF)
    • Background Calculation
    • Homozygous •Both are carriers
    • recessive disease •Use C for gene found on chromosome and F/f for the
    • 7 gene
    Answer Represent C F not carrying CF C f as carrying CF Your couple is heterozygous for CF
  • Example of sex linked traits:
    • Color Blindness
    • A sex linked gene
    • Carrier female x Normal Male
    • Hemophilia
            • Female carrier x Normal male
    Answer Answer X B X b X B y X H X h X H y
  • Example:
    • Sickle cell
    • Alleles are codominant .
    • Cross between two carriers with sickle trait.
  • Example Codominant
    • Petal Color 
    • Many alleles create 2 phenotypes from two alleles Codominant create 3 phenotypes from 2 alleles
    • As an example: the four o’clock petunea. One type has red petal (RR), another has white (WW) and a third has pink (RW). Pink are Heterozygotes
            • Cross Red X Red
            • Gene key is C
            • Red allele is a suffix of R
            • White allele is the suffix W
  • Pedigree Chart
    • Another way to visualize a monohybrid crosses or determining a genotype is by using a pedigree chart
  • Pedigree Chart
    • Another way to visualize a monohybrid crosses or determining a genotype is by using a pedigree chart
    • Knowing the phenotype of individuals in a family will sometimes allow genotypes to be determined.
    • In genetic counseling this enables probabilities to be determined for the inheritance of characteristics in children.
  • Pedigree Chart
    • White circle : Normal female
    • White Square: Normal male
    • Black Circle: affected female
    • Black square: affected male
    • (1) and (2)..Normal Parents
    • (3) affected female
    • (4),(5) and (6) normal
  • Pedigree Chart The chart shows the inheritance of the hemophilia gene through the Royal families of Europe
  • Study the diagram and the key then attempt the questions. Follow the link for the solutions.
    • Rupert had hemophilia what was his genotype?  
    • What is the genotype and phenotype of Alice, Rupert's mother?
    • What is the genotype and phenotype of Leopold, Rupert's grandfather?
    • Gonzalo and Alphonso are haemophiliacs and brothers. Who therefore is a carrier?
    • If Irene Princess Henry has sons can they be hemophiliac?
    • George the V did not have haemophilia. Can the disease be present in the current royal family of the United Kingdom?
    Answer
  • The following examples of pedigree relate to two human diseases:
    • 1.Phenylkentonuria(PKU)
    • is a genetic disease in which an
    • individual is unable to
    • Produce the enzyme PKU. This
    • enzyme changes the R group of
    • the amino acid Phenylalanine to
    • Tyrosine.
    • Answer the following
    • questions on this pedigree. 
  • 1. Phenylketonuria (Pku)
    • Using the allele key provided state the genotype of parents 1 and 2?
    • Give the genotype and phenotype of individual 5 ?
    • Is it possible that the condition is sex linked ?
    • What is the genotype and phenotype of individuals 7 and 8?
    • Which two individuals have the incorrect pedigree
    Answer
  • 2. Muscular Dystrophy
    • What type of genetic disease is muscular dystrophy?
    • Give the genotype and phenotype of 1?
    • Give the genotype and phenotype of 2?
    • Give the genotype and phenotype of 8 ?
    • Give the genotype and phenotype of 5 and 6 ?
    Answer
  • Topic 8.2: Dihybrid Crosses
  • Calculating and predicting genotypic and phenotypic ratios for unlinked autosomal dihybrid crosses
    • Terminology :
    • Dihybrid crosses involve two genes which control two characteristics. There are complications of these patterns as illustrated in the calculations that follow.
    • Unlinked genes are found on different chromosomes and can be segregated by random assortment of meiosis/ metaphase II
    • Autosomal are those chromosomes other than the XY gender determining chromosomes (not sex linked).
  • Mendel's Dihybrid cross with Peas
    • Phenotypes: Smooth Yellow Seeds X Rough Green Seeds
    • The chromosomes are shown as homologous pairs but not as bivalents.
    • Smooth is dominant to Rough
    • Yellow is dominant to green
    • Meiosis:
    • Reduces the chromosome number and randomly assorts the alleles for each gene
    • As the parents are homozygous they each produce only one type of gamete.
    • Fertilization:
    • Random fertilization of the gametes.
    • Offspring:
    • The offspring are heterozygous at each gene loci. Notice the chromosome number is restored to the that of the parents.
    • New homologous pairs are produced
    • The offspring are crossed = F1 x F1 (F1 self)
  • The grid shows all offspring genotype combinations from random fertilization
    • Phenotype Key:    
    • Phenotypic ratio:
      • 9 Smooth Yellow:
      • 3 Smooth green:
      • 3 Rough Yellow:
      • 1 Rough Green
                                          
  • Dihybrid Ratio :
    • This ratio is called the F2 Dihybrid Ratio and results from a cross between two heterozygous dihybrids
    • The ratio is a prediction of the offspring ratio.
    • Actual numbers may deviate from this ratio as each fertilization is a random process
    • The ratio only sets the probability of a particular offspring phenotype arising.
  • Dihybrid Cross Recombinant
    • Recombination occurs with the reassortment of genes or characters into different combinations from those of the parents. In the example below color blindness is carried only on the X chromosome Xb
    Allele Key: XB= Normal vision Xb= Colorblind Y = Male chromosome T= Tongue roller t = non-tongue roller Allele key: C= Color produced c = no color produced   A = Banding (agouti/back yellow tip) a =Non banding (black) Answer Answer XbXBTt X XByTt CcAa X CcAa
  • Example: Sweat Peas (Lathyrus odoratus)
    • This famous example comes from the work of W. Batson who rediscovered the work of Mendel and carried out his own experiments in the early 20 Century.
    • Allele Key:
    • Flower colour Purple (P) and Red (p)
    • Pollen grain shape, Long (L) and short (l)
    • A cross was made between a plants that where heterozygotes at both gene loci (PpLl).
  • Analysis of date:
    • There is a significant difference between observed data and expected data for linkage (3:1 ratio with no cross over)
    • The data do not agree with the theory )for an unlinked cross with a chi square calculation of P < 5%
    • Conclusion the genes are linked but some cross over has occurred.
  • 8.3.4 Linkage Group Crosses
    • In linkage crosses the format of the alleles
    • allows us to see which alleles are inherited
    • together and which are recombinants.
    • Using the example of Sweet Peas the genes for Flower color and pollen grain shape can be shown. This shows the linkage of the P allele with the L allele with the line representing the chromosome.
    • The heterozygote (at both loci)
    • would be represented as follows
    • During metaphase II recombinant would show the chromosome after cross over has completed. In this case there could be the following recombinant alleles in the gamete.
    OR
  • Example: Drosophila (Fruit Fly)
    • Normal wing (L) is linked to Grey Body (G) whilst Vestigial wing (l) is linked to the mutant black (b) body.
    • A test cross of a potential heterozygote produced the ratio of
    • Grey Normal (283),Grey Vestigial(1294),
    • Black Normal(1418) and Black Vestigial (241)
    • Using Chi square test determine if the genes are linked or unlinked.
  • Statistical tests for genetic crosses (Chi Square)
    • Sometimes the data from an experiment are not measurements but counts (or frequencies) of things, such as counts of different phenotypes in a genetics cross.
    • With frequency data you can’t usually calculate averages or do a t test, but instead you do a chi-squared test.
    • This compares observed counts with some expected counts and tells you the probability (P) that there is no difference between them.
    • In Excel the the test is performed using the formula: =CHITEST (observed range, expected range) .
    • There are three different uses of the test depending on how the expected data are calculated. Chart link will show you why we choose the test below.
  • (Chi Square)
    • Sometimes the expected data can be calculated from a quantitative theory, in which case you are testing whether your observed data agree with the theory.
    • If P < 5% then the data do not agree with the theory.
    • If P > 5% then the data do agree with the theory.
  • Example: Sweet Peas
    • Allele Key:
    • Flower color Purple (P) and Red (p)
    • Pollen grain shape, Long (L) and short (l)
    • A cross was made between a plants that where
    • heterozygotes at both gene loci (PpLl). Only one
    • shows cross over in gamete genotypes.
    • Show the possible genotypes and identify
    • recombinants. Assume P is linked to L
  • Chi square calculation on dihybrid data
    • In one of Mendel's dihybrid crosses, the following types and numbers of pea plants were recorded in the F2 generation:
    • Yellow round seeds (289)Yellow wrinkled seeds(122) Green round seeds(96) Green wrinkled seeds(36)
    • According to theory these should be in the ratio of 9:3:3:1. Do these observed results agree with the expected ratio?
  • Chi square calculation on dihybrid data
    • In one of Mendel's dihybrid crosses, the following types and numbers of pea plants were recorded in the F2 generation:
    • Yellow round seeds (289)Yellow wrinkled seeds(122) Green round seeds(96) Green wrinkled seeds(36)
    • According to theory these should be in the ratio of 9:3:3:1. Do these observed results agree with the expected ratio?
    Answer
  • Topic 8.4 Polygenic Inheritance
  • Polygenic Inheritance (multi alleles)
    • Definition: 'A single characteristic that is controlled by two or more genes‘
    • Each allele of a polygenic character often contributes only a small amount to the over all phenotype. This makes studying the individual alleles difficult.
    • In addition environmental effects smooth out the genotypic variation to give continuous distribution curves.
  • 8.4.2 Polygenic characteristic more often show continuous variation
    • (a) Is the genotypic variation in the population
    • (b) Phenotypic variation = genotypic variation + environmental variation.
    • Vp = Vg + Ve
  • 3.3.5 ABO Blood Groups
    • An example of multiple alleles is blood groups with 3 alleles
  • Multiple Alleles: ABO Blood Groups Blood type O: Universal donor. Blood type AB: Universal acceptor
  •  
  • Example 2: Polygenic Inheritance/ Human skin color
    • This is controlled by as many as 4 genes each with its own alleles.
    • As the number of genes increases the amount of phenotypic variation increases.
    • The alleles control the production of melanin which is a pigment that colors skin.
    • In this example the calculation is performed with 2 genes each with 2 alleles. The cross is between two individuals heterozygous at both alleles
    Allele Key A= add melanin a= no melanin added B= adds melanin b= no melanin added
    • The graph shows the variation in skin color and frequency. if the calculation is performed for 4 genes the graph smoothes out the different color types
    • The frequency of groups will be affected by the local gene pool and this graph should not be over interpreted for skin group frequency
  • Example 3: Finch Beak Depth
    • Finches are seed eating birds that use their beaks to break open seeds. The depth of beak is under polygenic control of three genes with two alleles each.
    • Produce the Punnett square for the following cross.
    • Compare the data frequency against the graphic representations to the right.
  • Example 4
    • Cross: between two bird which are heterozygous at
    • All three loci
          • Allele key :
          • A= add depth
          • a= no depth added
          • B= add depth
          • b= no depth added
          • C= add depth
          • c= no depth added
    Answer
  • Example: Wheat Color Wheat polygenics was first worked out in 1909 by Nilsson- Ehile
    • Each of the alleles (for the three genes) of the wheat has a small effect on the phenotypic variation for color. The additive effect produces the continuous variation that is found in Wheat red coloration
  • Example: Wheat Color
    • Cross :
    • Two plants heterozygous at each 3 gene loci
    • calculate the phenotypic ratio and draw a graph of the phenotypic variation
    • Allele Key :
    • 3 genes with 2 alleles each. A,B and C
    • Each 'dominant' allele produces one unit of color.
    • The homozygous dominant at all three loci produces dark red wheat.
    • The homozygous recessive at all three loci produces pale white color.
    Answer
  • Example 5 : Wild Columbine
    • Another example of how the height of a plant is controlled by a polygenic system. Here there are two genes each with two alleles. Again cross two heterozygote's at each loci.  
        • Allele key :
        • M1= Tall plant
        • M2= Small plant
        • N1=Tall plant
        • N2=Small plant
    Answer
  • Example 6: Chicken Combs
    • Controlled by two genes this produces four phenotypes. Using the allele key:
    • gene 1: P and p
    • gene 2: R and r
    • Exercise :
    • Produce a completed
    • allele key and a punnett
    • grid with genotypic and
    • phenotypic ratios that
    • explain the phenotypes
    • to the right
  • Practice
    • Dihybrid Calculations   
    •   Example 1 : The Capercaille is a ground dwelling bird of the pine forest of Scotland (simple dominance/ recessive). The length and color of the primary feather is controlled by two unlinked genes. Using the allele key to the right calculate the phenotypic ratio of an F2 beginning with a cross between a hen bird that is homozygous recessive for both genes and a cock bird that is homozygous dominant for both genes.
    Allele Key: Q- Long feather q-short feather A- dark feather color a-light feather color
  • Example2: Sweet peas (codominance/ dominance)
    • Sweet pea flowers color is controlled by a codominant pair of alleles. The flowers are white, pink and red. The height of the mature plant is controlled by a gene showing dominance. Calculate the F2 phenotypic ratio for a cross between two heterozygous plants at both gene loci.
    Allele key: A1= white, A2= red B= Tall, b= short Answer
  • Example 3: Dihybrid Test Cross
    • A suspect heterozygote Guinea pig for coat color and length is crossed with a double homozygote recessive. As with a monohybrid test cross there is a predictable dihybrid test cross ratio which is 1:1:1:1
    Allele key: L = long hair coat l = short hair coat B = Black coat b =brown coat Answer
  • Answer to capercaille * Practice
  • Sweat Pea Example Back
  • Guinea Pig Back
  • Example 4
  • Sweet Pea Example
  • chi Back
  • Example Back
  • Example: Wheat Color Back
  • Example: Wild Columbine Back
  • Cystic fibrosis Back to slide
  • Color Blindness Back to slide
  • Hemophilia
  • Does Rupert have hemophilia Return
  • PKU answer Back
  • Muscular Dystrophy (MS) Back
  • Tongue Roller Back
  • Mice Back
  • Sickle Cell Cross