GENETICS - CLUTCH CH.3 EXTENSIONS TO MENDELIAN INHERITANCE.

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2 CONCEPT: INDEPENDENT ASSORTMENT Independent Assortment is Mendel s Second Law Independent Assortment: States that alleles of two genes assort independently A dihybrid cross is used to examine two independently assorting genes EXAMPLE: Genotype Phenotype AaBb Yellow Round What are the genotypes of the gametes? Page 2

3 PRACTICE 1. Which of the following gametes cannot be formed from the genotype AaBBCc? a. abc b. ABC c. Abc d. ABc 2. Which of the following gametes cannot be formed from the genotype DDeeFfGG? a. DeFG b. DDeFG c. DefG d. DefG Page 3

4 3. Which of the following gametes cannot be formed from the genotype HhJjKK? a. HJK b. hjk c. hjk d. HjK Page 4

5 CONCEPT: CHI-SQUARE ANAYLSIS A chi-square test is used to examine whether to expected result is closet enough to the observed result Genetics is never you wont get a perfect 3:1 ratio or 9:3:3:1 ratio - A chi-square test is used to check if your numbers are close enough to expected ratio - Observed numbers The numbers you actually get - Expected numbers The numbers that you expected to get. The perfect ratio numbers Formula: - Observed = o - Expected = e PRACTICE : A. You bred a purple plant, which you think is heterozygous (Aa), with a homozygous recessive (aa) white plant. There were 120 offspring produces, 55 are purple and 65 are white. Was your red plant heterozygous? Assume Mendelian inheritance. 1. Determine what the expected ratio of the heterozygous (Aa) cross with a homozygous recessive (aa) plants would be? - If there are 120 offspring would be purple and would be white. - These are your expected numbers Page 5

6 2. Use tuse the chi-square formula to calculate your chi square value CLASS O E (O-E) 2 (O-E) 2 / E Red White Total Use the chi-square distribution table to determine whether or not your hypothesis is true. a. Calculate your degrees of freedom; df = # of variables - 1 i. There are two variables in this problem (purple and white) ii. Df = 2-1 = 1 b. Within your calculated degrees of freedom row, find where your chi-square value would be. c. Determine the range of probability (p value) i. For this problem is is , which is 50%-30% Page 6

7 4. Determine If you accept or reject your null hypothesis a. The null hypothesis states there is no difference between measured and predicted values i. In this problem, the null hypothesis would be that 55 red and 65 white plants is close enough to 60 red and 60 white plants and therefore they are not different. b. Generally, you accept (fail to reject) the null hypothesis if the probability is greater than 5% or 0.05 c. Generally, you reject the null hypothesis if the probability is less than 5% or The probability for this question was between 30% and 50% (0.3 and 0.5) therefore we accept the null hypothesis 5. Figure out what the null hypothesis means for our actual problem a. Accepting the null hypothesis, means that the observed and expected aren t different b. Therefore, we are 95% confident that the purple plant was heterozygous Page 7

8 PRACTICE Using the following F2 phenotypes from a monohybrid cross, answer the following question. F2 Phenotype # of F2 Offspring Red Flowers 892 White Flower Which of the following null hypothesis is the best to test using the chi-square test? a. There is no difference between my values and an expected 3:1 ratio b. There is no difference between my values and an expected 2:2 ratio c. There is no difference between my values and an expected 9:3:3:1 ratio d. There is no difference between my values and an expected 3:2 ratio Using the following F2 phenotypes from a monohybrid cross, answer the following question. F2 Phenotype # of F2 Offspring Red Flowers 892 White Flower Which of the following represents the appropriate degrees of freedom for this problem? a. 1 b. 2 c. 3 d. 4 Page 8

9 Using the following F2 phenotypes from a monohybrid cross, answer the following question. F2 Phenotype # of F2 Offspring Red Flowers 892 White Flower Using the chi-square formula, calculate the chi-square value. a b c d Assuming a chi-square value of and a single degree of freedom, what is the range of p-values? a b c d Page 9

10 5. If a chi-square value has led you to receive a p-value range of , will you accept or reject the null hypothesis? a. Accept the null hypothesis b. Reject the null hypothesis 6. Which of the follow statements is true when we accept a null hypothesis. a. The observed and expected values are different b. We are 95% confident that our observed and expected values are different c. We are 95% confident that our observed and expected values are the same d. We are 50% confident that our observed and expected values are the same Page 10

11 CONCEPT: SEX CHROMOSOMES Sex Determination Different organisms determine sex differently There are two types of sex s - Homogametic sex is the sex that has only one type of chromosome (X) - Heterogametic sex is the sex that has two types of chromosomes (X or Y) There are three different sex systems - XX/XO: Sex s are determined by having either one or two of the same chromosome - XX/XY: Sex s are determined by having either one or two types of chromosomes - ZZ/ZW: The male is homogametic and the female is heterogametic Organisms can be - Monoecious: when an organism contains both male and female sexual organs (hermaphrodite) - Dioecious: when an organism contains either male or female sexual organs EXAMPLE: Sex determination systems Humans: XX:XY Insects: XX:XO Birds:ZW:ZZ Female Gametes X X X X Z W X XX XX X XX XX Z ZZ ZW Y XY XY - X- X- Z ZZ ZW Male Gametes Male Gametes Male Gametes Females: XX or ZW Males: XY or XO or ZZ Page 11

12 Human Sex Chromosomes Humans follow the XX/XY model, with males determined by the XY chromosomes During meiosis, the XY or XX chromosomes, separate into the haploid daughter cells - Nondisjunction is when chromosomes do not separate properly during meiosis There are many that can occur if the sex chromosomes do not segregate properly - Klinefelter syndrome (47 XXY) : have male genitalia, but usually fail to produce sperm - In severe cases, klinefelters can have more Xs and display more severe phenotypes - Turner syndrome (45 X): have female genitalia, but have cognitive and sexual impairments - Mosaics can occur when somatic cells display two different numbers of sex chromosomes - Example: 45X / 46 XY EXAMPLE: Karyotype for Klinefelter Syndrome Page 12

13 Human sex s handle their chromosomes differently The Y chromosome determines - Sex determining region (SRY) produces a testis determining factor (TDF) protein Females get two of the X chromosome - Therefore one needs to be inactivated so females don t get double the dose of X they need - Barr bodies are the name for the inactivated X chromosome - It is inactivated by condensing the chromatin - The lyon hypothesis states that X chromosome inactivation occurs randomly in somatic cells - Occurs early in development, so it can result in a mosaic phenotype Ex: calico cats EXAMPLE: Calico cats and the Lyon Hypothesis Page 13

14 PRACTICE 1. Which of the follow sex chromosomes can be describes as homogametic? a. XX/XY b. ZZ/ZW c. XX/XO 2. In the ZZ/ZW sex determination system, the male is? a. Heterogametic b. Homogametic c. ZW d. XY Page 14

15 3. In humans, which region on the Y chromosome determines maleness? a. Sperm producing factor (SPF) b. Testis determining region (TDR) c. Sex Determining Region (SRY) d. Barr Body (BB) 4. During meiosis the XY sex chromosomes segregate independently. Which of the following represents the gametes chromosomes after meiosis? a. Two haploid cells that each contain X or Y b. Two diploid cells that both contain an XY c. A single diploid cell that contains XY d. Four haploid cells two contain X and two contain Y Page 15

16 CONCEPT: X INACTIVATION Dosage compensation is a phenomenon where the gene expression of sex chromosomes is similar in both sexes EXAMPLE: Dosage compensation makes up for the fact that different sexes have chromosomal numbers X-inactivation is the shut-down of on x chromosome (forms barr body) - X-inactivation center (Xic) on x chromosome that is required for inactivation - Xist gene produces an RNA molecule that coats the X chromosome and inactivates it - Tsix gene prevents X inactivation - Xist RNA molecule helps to recruit proteins to the X chromosome which form a Barr body Barr Body Xist Page 16

17 PRACTICE: 1. Why must one of the X chromosomes in human females undergo X-inactivation? a. Because all X chromosome alleles are dominant b. Because of dosage compensation c. Because all X chromosome alleles are recessive d. Because the X chromosome is not needed for normal development 2. What is NOT a region on the X chromosome required for X-inactivation? a. Xic b. Xist c. Tsix Page 17

18 CONCEPT: OVERVIEW OF INTERACTING GENES Genetics is very rarely clear cut, as many traits exist on a (ex: height or weight) Polygenic describes traits controlled through multiple traits - Often the expression is continuous (ex: height) EXAMPLE: Assume there are two genes (P1 and P2) that both determine a plant s color (purple). P1/p1/P2/p2 x P1/p1/P2/p2 You get four offspring outcomes 1. Homozygous or Heterozygous dominant for both genes P1/-/ P2/- 2 doses 2. Homozygous or heterozygous dominant for gene 1 P1/-/ p2/- 1 dose 3. Homozygous or heterozygous dominant for gene 2 p1/-/ P2/- 1 dose 4. Homozygous recessive for both gene 1 and gene 2 p1/-/ p2/- 0 doses Multiple genes can interact, causing very different phenotypic effects - Pleiotropy: describes when a single gene has multiple effects on the phenotype of an organism - Variations of Dominance: describes multiple ways a dominant allele can effect the phenotype - Epistasis: is the interaction of two different genes and how the interaction effects the phenotype - Penetrance: is the percentage of individuals with a given allele who exhibit the phenotype - Expressivity: measures the degree to which a given allele is expressed at a phenotypic level Page 18

19 PRACTICE: 1. Which of the following terms describes an interaction between two genes? a. Penetrance b. Pleiotropy c. Incomplete dominance d. Epistasis 2. Polygenic traits are controlled through which of the following ways. a. Multiple alleles for one gene b. Multiple genes c. Epistasis d. Penetrance Page 19

20 3. True or false: Polygenic traits are usually continuous traits? a. True b. False Page 20

21 CONCEPT: PLEIOTROPY Pleiotropy describes when a single gene has multiple effects on the phenotype of an organism Pleiotropy occurs in main ways 1. Expression of a single gene can have multiple functions 2. Gene is expressed in different cell types in multicellular organisms 3. Gene is expressed at differential stages of development EXAMPLE: Cystic fibrosis CFTR protein CFTR is a transmembrane protein that balances Cl - concentrations - When mutated, it can no longer do this properly - Results in thick mucus in lungs, and excessively salty sweat, among other issues Page 21

22 PRACTICE 1. Which of the following is NOT an example of a pleiotropic trait? a. A single gene is expressed and has different functions in the brain, arm, and kidneys b. A single gene can increase blood pressure, fire neurons, and keeps mucus in the lungs c. Two genes are responsible for filtering urine in the kidneys d. A single gene is expressed consistently throughout the life of an organism, but as the organism changes it has different effects 2. A pleotropic trait is controlled by a. 1 gene b. 2 genes c. 3 genes d. 4+ genes Page 22

23 CONCEPT: EPISTASIS AND COMPLEMENTATION Complementation A complementation test is performed to determine if two mutants have mutations in the same gene A complementation test is performed by mating two recessive mutants with the same phenotype together - If offspring are WT: The two mutations are in different genes - If offspring are mutant: The two mutations are in the same gene EXAMPLE: You have three white mutants (1,2,3) and want to know if the mutation causing them each to be white is in the same gene for each mutant. WT color is blue. Which mutations complement (meaning the two mutations are in different genes)? White 1 x White 2 F 1 all white White 1 x White 3 F 1 all blue White 2 x White 3 F 1 all blue Page 23

24 Non-Epistatic Genes Epistasis is the interaction of two different genes - Generally, the presence of one gene s allele will the phenotype of the second gene s allele - Two non-epistatic genes will show the expected 9:3:3:1 dihybrid ratio - Genes under epistasis will show an altered ratio EXAMPLE: Corn snakes come in four color patterns: orange, black, camouflaged, and albino. - The color pattern is determined by two different genes (O,o or B, b). Genotype O + /- ; b/b o/o ; B + /- O + /- ; B + /- o/o ; b/b Phenotype Orange Black Camouflaged Albino What is the offspring s genotype and phenotype derived from the mating of two heterozygous camouflaged corn snakes? O + /o ; B + /b x O + /o ; B + /b Orange Black Camouflaged Not black Orange Not Orange Black Black Not Black Albino Page 24

25 Epistatic Genes Dominant epistasis occurs when a dominant allele of one gene masks the effects of either allele of the 2 nd gene EXAMPLE: - The dominant allele is epistatic - The offspring phenotypic ratio of a heterozygous cross is 12:3:1 A certain breed of squash comes in three colors; white, dark red, and light red. Coloration is determined by two genes, D and W. Genotype Phenotype Genotypic Ratio Phenotypic Ratio D/- ; W/- White 9 12 d/d ; W/- White 3 D/- ; w/w Dark red 3 3 d/d ; w/w Light red 1 1 Phenotypic ratio is 12:3:1 because the dominant W allele is epistatic. - This means that when there is a dominant W present in the offspring, it will mask the phenotype of the D allele EXAMPLE: Recessive epistasis occurs when a recessive allele of a gene masks the effects of either allele of the 2 nd gene - The offspring phenotypic ratio of a heterozygous cross is 9:3:4 A certain breed of flower comes in three colors: blue, magenta, and white. Coloration is determined by two genes, w + and m +. Genotype Phenotype Genotypic Ratio Phenotypic Ratio w + /- ; m + /- Blue 9 9 w + / ; m/m Magenta 3 3 w/w ; m + /- White 3 4 w/w ; m/m White 1 Phenotypic ratio is 9:3:4 because the mutant w allele is epistatic - This means that when an organism has w/w it will mask the phenotype of the m + or m alleles. Page 25

26 EXAMPLE: Rare example of recessive epistasis in humans: The Bombay Phenotype There are two genes responsible for blood type those in the I family, and those in the H family - The I family determines what protein will be on the blood cells (A, B, both, or none) - The dominant H allele makes a protein that adds the proteins onto the blood cell Genotype Blood Type Protein I A /I A or I A /i A A I B /I B or I B /i B B I A /I B AB AB i/i O none There is a rare mutation, represented by h, which doesn t create the protein needed to attach the enzyme - The h/h genotype is epistatic, meaning that it masks the phenotype of any I allele Genotype I A /I A or I A /I and hh I B /I B or I B /I and hh I A /I B and hh i/i and hh Blood Type O O O O Page 26

27 Other Gene Interactions EXAMPLE: Complementary gene action occurs when the two genes interact because they re in the same pathway - The offspring phenotypic ratio of a heterozygous cross is 9:7 A breed of flower comes in two colors; purple and white. Coloration is determined by two genes, C and P Genotype Phenotype Genotypic Ratio Phenotypic Ratio C/- ; P/- Purple 9 9 C/- ; p/p White 3 7 c/c ; P/p White 3 c/c ; p/p White 1 - The genes are complementary, because both genes need to have a dominant allele in order to be purple. EXAMPLE: Suppressors are mutant alleles of one gene that reverses the effect of a mutation in a second gene - When the suppressor causes the phenotype to be like WT, then the F2 ratio is 13:3 - When the suppressor causes the phenotype to be mutant, then the F2 ratio is 14:2 A breed of flower comes in two colors; The WT red and the mutant purple color. Coloration is determined by two genes p + and R. Genotype Phenotype Genotypic Ratio Phenotypic Ratio p + /- ; R/- Red 9 12 p + /- ; r/r Red 3 p/p ; R/- Purple 3 3 p/p ; r/r Red 1 1 (total 13) - The p + allele and the R allele both cause the plant to be red. - The mutant p allele causes the plant to be purple, UNLESS the recessive suppressor (r/r) is present, which cancels or reverses the purple phenotype and turns the plant red. Page 27

28 EXAMPLE: Modifiers occur when a mutation at one gene changes the degree of expression of a mutated second gene Genotype Protein phenotype a + /- ; b + /- WT a + / ; b/b Defective (low transcription) a/a ; b + /- Defective (mutated protein A) a/a ; b/b Extremely defected EXAMPLE: Synthetic lethal alleles occur when two viable single mutations result in death when found as double mutants - F2 generation is 9:3:3 Genotype Phenotype Genotypic Ratio Genotypic Ratio C/- ; P/- Purple 9 9 C/- ; p/p Cyan 3 3 c/c ; P/p White 3 3 c/c ; p/p Dead 1 Dead Page 28

29 PRACTICE: 1. When performing a complementation test, how do you know if two mutations complement? a. The offspring will have a wild-type phenotype b. The offspring will have the mutant phenotype c. The offspring will have an intermediate phenotype between wild-type and mutant d. The offspring will not look like either wild-type or mutant 2. How can you tell if two genes are epistatic? a. The F2 offspring from a cross show a 9:3:3:1 phenotypic ratio b. The F2 offspring all show a mutant phenotype c. The F2 offspring show a phenotypic ratio that is NOT 9:3:3:1 d. The F2 offspring all show a wild-type phenotype Page 29

30 3. Two heterozygous organisms are crossed, and the F2 phenotypic ratio is 12:3:1. What form of epistasis do these two genes exhibit? a. Dominant Epistasis b. Recessive Epistasis c. Suppressors d. Complementary Gene Action 4. A cross of white plants and red plants was performed. Using the F2 phenotypic ratio data below, determine what form of gene interaction is taking place. Phenotype # of offspring Red 1125 White 875 a. Dominant Epistasis b. Recessive Epistasis c. Suppressors d. Complementary Gene Action Page 30

31 5. In the rare Bombay phenotype, a mutation in a second gene can control an individual s what? a. Hair color b. Height c. Blood type d. Skin color Page 31

32 CONCEPT: VARIATIONS ON DOMINANCE There are many types of Complete dominance is when the dominant allele is always expressed when at least one copy is present - Homozygous dominant is not phenotypically different from heterozygous EXAMPLE: R R r Rr Rr r Rr R Rr R r Rr Rr Incomplete dominance is when the number of dominant alleles affects the phenotype - Homozygous dominant appears different r than heterozygous Rr Rr EXAMPLE: What is the phenotype of the offspring from a red Rflower mating and white flower mating? R r Rr Rr r R Rr R Rr r Rr Rr r Rr Rr Page 32

33 Codominance is when there are two dominant alleles, and both are expressed - Ex: ABO Blood Types EXAMPLE: Genotype I A /I A or I A /i I B /I B or I B /i I A /I B i/i Blood Type A B AB O I B i I A Type AB I A I B Type A I A i i Type B I B i Type O ii PRACTICE: 1. Which of the following is NOT a type of dominance? a. Complete dominance b. Incomplete dominance c. Codominance d. Leaky dominance 2. Which of the following parents could produce offspring with an AB blood type? a. A x A b. O x A c. A x AB d. AB x O Page 33

34 3. Blood types are an example of what type of dominance? a. Complete dominance b. Incomplete dominance c. Codominance d. Leaky dominance Page 34

35 CONCEPT: PENETRANCE AND EXPRESSIVITY Organisms sometimes have a genotype that doesn t completely their phenotype There are three reasons a genotype wouldn t give the expected phenotype 1. Environmental factors influence or mask the phenotype 2. Influence of other interacting genes (epistasis, modifiers, suppressors) 3. The phenotype is subtle and difficult to observe Penetrance is the percentage of individuals with a given allele who exhibit the phenotype Expressivity measures the degree to which a given allele is expressed at a phenotypic level - Ex: Brown fur can have different intensities of brown (light to dark) EXAMPLE: Penetrance vs. Expressivity Penetrance Expressivity Page 35

36 PRACTICE: 1. A group of individuals exhibit a range of intermediate phenotype between dominant and recessive. What term measures this phenotype? a. Incomplete dominance b. Penetrance c. Expressivity d. Epistasis 2. Penetrance measures which of the following? a. The degree to which an allele is expressed b. The effect of two genes on a single trait c. A single intermediate phenotype between dominant and recessive d. The percentage of individuals with a given allele who exhibit the phenotype Page 36

37 CONCEPT: ORGANELLE DNA Mitochondria and chloroplasts both contain their own DNA (mtdna and cpdna) Cells are classified based on the of DNA sources they contains - Heteroplasmic cells contain DNA in the nucleus, and from organelle sources - Homoplasmic cells contains only DNA from one source The endosymbiont theory explains how mitochondria and chloroplasts evolved with eukaryotic cells - It states that the mito and chloro were free-living bacteria, that were engulfed by early eukaryotic cells Mutations in the organelle DNA can cause serious defects - Myoclonic epilepsy and ragged-red fiber disease (MERRF) causes deafness, seizures, and other issues EXAMPLE: Endosymbiont Theory The DNA found in these organelles is small, and - It generally contains few noncoding regions, and no introns Human mitochondrial DNA has certain properties - It has a heavy chain, which has more guanines - It also has a light chain, which has more cytosines The codon code is not universal in mitochondria and chloroplasts - Ex: AGA codes for arginine normally, but in Drosophilia mitochondrial DNA it codes for serine Page 37

38 Inheritance Inheritance of organelle DNA is different than DNA Uniparental inheritance is when progeny inherit DNA solely from one parent - Mitochondrial DNA is inheritance maternally EXAMPLE: Mutant Female X WT Male Progeny all mutant WT Female X Mutant Male Progeny all WT Cytoplasmic segregation is when two organelles apportion themselves into different daughter cells - Ex: Variegation in plants, slow growing poky Neurospora fungi - In rare cases, an affected female will produce a healthy offspring EXAMPLE: Variegation due to cytoplasmic segregation of chloroplasts in plants Page 38

39 CONCEPT: MATERNAL EFFECT GENES Maternal effect occurs when the offspring s phenotype is determined by present in the egg itself Linnaea shell coiling is an example of a maternal effect genes - DD and Dd genotypes form right-handed coiled shells - dd genotypes form left-handed coiled shells - A dd offspring will form a right-handed coiled shell if the mother is DD or Dd EXAMPLE: Page 39

40 PRACTICE: 1. What controls the inheritance of an offspring s phenotype when the trait is controlled through the maternal effect? a. The offspring s genotype b. The mother s genotype c. The father s genotype 2. True or False: Maternal inheritance and maternal effect are two terms for the same type of inheritance. a. True b. False Page 40