Genetics Notes

Genetics Study Guide Chapter 3 & 4: A. Mendel’s 4 postulates • 1. Alleles: -there are 2 fms. of a gene, dominant & recessive; dom. will hides recessive and recessive will only express itself if paired w/another recessive • 2. Law of Segregation: -members of each pair of alleles separate when gametes are formed. A gamete will receive one allele or the other -genotype: TT/ Tt, these are the 2 alleles one has for specific gene -phenotype: the appearance of the trait (ex: tongue rolling) • 3. Homozygous vs. Heterozygous -homozygous: if a person has 2 alleles that are alike (TT or tt) -heterozygous: 2 alleles that differ (Tt) -homozygous dominant: TT -homozygous recessive: tt -monohybrid cross: looks at one contrasting trait, mate 2 individuals (parents=P); first look at the first offspring generation (F1), self cross w/ this would be F2; ratios will be 1:2:1 genotypic ratio and 3:1 phenotypic ratio -test cross: don’t know the genotype so cross it w/ homozygous recessive individual to find out -dihybrid cross: RrYy X RrYy, ratio of 9:3:3:1 • 4. Law of Independent Assortment: -alleles for different traits are distributed to sex sells and offspring independently of one another

B. Vocab. Chapter 3 & 4 1. gene interaction- a situation in which a single phenotype is affected by more than 1 gene 2. X-linkage- cases where genes are present only on the X-chromosome 3. wild-type allele- alternative forms of the gene in organisms where this allele occurs most frequently; usually dominant

4. incomplete dominance-(partial dominance) a cross between 2 alleles produces an 5. 6. 7. 8. 9. 10. 11. 12. 13.

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intermediate result (ex: when red flowers cross w/white flowers = pink) ratio is 1:2:1 Tay Sachs disease- human biochemical disorder where homozygous recess. individs. severely affected w/ fatal lipid-storage disorder causing babies to die hexosaminidase- involved in lipid metabolism, in Tay Sachs there is no activity in this codominance- joint expression of 2 alleles in heterozygote (ex: MN blood group in ppl.) ABO Blood groups-simplest case of multiple alleles 3 alternative alleles of 1 gene exits isoagglutinogen- the “I” in blood groups, also an antigen H substance- something nearly everyone has where terminal sugars are added secretor locus- expression of the ABO blood type system is affected by this; are the antigens present in secretions (blood, saliva)? Rh antigens- another set of antigens illustrates multiple allelism; involved in erythroblastosis fetalis, form of anemia recessive lethal allele- mutations resulting in a gene product that is nonfunctional can sometimes be tolerated in the heterozygous state; one wild type might produce enough of the essential product for the organism to survive. If it behaves recessively, the organism will not survive if it is homozygous recessive dominant lethal allele- in instances where 1 copy of the wild-type gene is not sufficient for normal development, even the heterozygote will not survive (ex: Huntington’s disease) epigenesis- a situation (such as development of the insect eye which is exceedingly complex and leads to a structure w/multiple phenotypic manifestations) where

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each succeeding step of development increases complexity of the sensory organ and is under ctrl. & influence of many genes epistatic- when the homozygous presence of a recessive allele may prevent/ override the expression of other alleles at a second locus; when studying the characteristic a ratio expressed in 16 parts (3:6:3:4) suggests epistasis recessive: occurs when the homozygous recessive genotype masks/ suppresses expression of another gene (Bombay phenotype for ABO blood groups) dominant: dom. allele at 1 gen. locus masks expression of allele at another locus (summer squash colors) hypostatic- the masked alleles at the second locus complementation analysis- helps determine whether 2 genes that are involved in a particular thing (formation of wings) are on the same gene, whether they are alleles or separate genes (sometimes called cis-trans-test, discovered/ created by Edward B. Lewis) complementation group- all mutations determined to be present in any single gene are said to fall into the same complementation grp. hemizygous- b/c males can’t exhibit either homozygous or heterozygosity for X-linked genes, this is called hemizygous criss cross pattern of inheritance- where phenotypic traits ctrld. by recess. X-linked genes are passed form homozyg. mothers to sons. Occurs w/females exhibiting recess. trait chromosome theory of inheritance-behavior of chromosomes during meiosis: fact that genes are transmitted on specific chromosomes sex limited inheritance- when expression of specific phenotype is absolutely ltd. to 1 sex sex-influenced inheritance- the sex of an individual influences the expression of a phenotype, but is not limited to 1 sex penetrance- % of individuals that show at least some degree of expression of a mutant genotype expressivity- the range of expression of the mutant genotype genetic suppression- the expression of genes throughout the genome have an effect on the phenotype produced by the gene in question; when 1 gene suppress the expression of another position effect- the physical location of a gene in relation to other genetic material may influence its expression heterochromatin- when the gene is relocated to or near certain areas of the chromosome that are prematurely condensed and genetically inert conditional- mutations that are affected by temperature are “conditional” & “temperature sensitive” (Himalayan rabbits & Siamese cats’ coat color changes) nutritional mutations- when the phenotype is not a direct reflection of the organism’s genotype auxotroph- when an enzyme essential to biosynthetic pathway becomes inactive (ex: phenylketonuria (PKU) cannot metabolize galactosemia genetic anticipation- studying the genetic onset of phenotypic expression has intensified the discovery of heritable disorders that exhibit a progressively earlier age of onset and an increased severity of the disorder in each successive generation genomic (parental) imprinting- variation of phenotypic expression depending strictly on the parental origin of the chromosome carrying a particular gene; in some species certain chromosomal regions and the genes contained w/in them somehow retain a memory or an imprint of their parental origin that influences whether specific genes either are expressed or remain genetically silent. Prader-Willi syndrome- results when only an undeleted maternal chromosome remains Angelman syndorm (AS)- when only an undeleted paternal chromosome remains

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37. Pleiotropy- expression of a single gene has multiple phenotypic effects, v. common among human genetic disorders (ex: Marfan Syndrome, heart problems, long fingers, barrel chested, very tall.) C. Conditional Prob. 1. asking what is the probability that one outcome will occur give certain conditions 2. Pc = Pa / Pb (Pa chance of 1 dominate, 1 recessive; Pb prob. of phenotype) -ex: parents are carriers of CF, what is a child’s chance of being a carrier -mom is Cc dad is Cc, Punnett square reveals ½ chance divided by ¾ + 2/3 chance (chance of heterozygous divided by total genotype) D. Binomial Therom - -Birth order problems 1. one of many alternative outcomes is possible 2. use it to calculate the probability of any specific set of outcomes among a lrg. # of potential events 3. ex 1: 7 kids, what is prob that ALL will be male: -(a + b) ^n= 1, use Pascal’s triangle (to the right) 1.) (a + b) ^0 =1 2.) (a + b) ^1 =a + b 3.) (a + b) ^ 2 = a^2 + 2ab + b^3 4.) (a + b) ^3= a^3 + 3a^2 b + 3ab^2 + b^3 -a & b are the respective probabilities of 2 alternative forms, n= # of trials - a= # boys, 7 & b= #girls, 0 - # trails= 7, so chose the 8th row (1st one doesn’t count) - 1a^7; 7a^6 b; 21a^5 b^2; 35a^4 b^3; 35a^3 b^4; 21a^2 b^5; 7a b^6; 1b^7 -chose the one w/ all as (boys) 1b^ 7 -P = 1 * (1/2)^7 = 1/128 -ex 2: What is probability of a family of 4 having AT LEAST 1 girl -simply add up all that have 1 girl -(a + b) ^4= 1a^4 + 4a^3 b + 6a^2 b^2 + 4ab^2+ 1b^4 - plug in (1/2) for each a and b, then add them up = 15/16 chance -ex 3: What is probability of having 2 boys, 2 girls in family of 4? -a= # boys, 2 b=# girls, 2 trials: n=2 -chose the term that has 2 boys and 2 girls -(a + b) ^4= 1a^4 + 4a^3 b + 6a^2 b^2 + 4ab^2+ 1b^4 -plug in the (1/2) for the a’s and b’s - 6 (1/2)^2 * (1/2)^2 - = 6/16 or 3/8, so in families w/ 4 kids, 3 out of 8 should be 2 girls, 2 boyts D. Summative and Additive law (dice questions) 1. Genetic event prediction: -predict the outcome of each fertilization even in the term of genetic potential (ex: ½ short plants, ½ tall) 2. Product law: - 2 or more events occurring independently of one another, but at the same time (AND), used when calculating probabilities via forked line method - ex: what is the chance that you will roll snakes eyes w/ 2 dice. chance of rolling 1 w/ first die = 1/6 chance of rolling 1 w/ 2nd =1/6 chance of rolling 2 1’s = 1/6 * 1/6= 1/36 *(rolling one does not effect outcome of 2nd role)* - genetic ex: what is prob of green, wrinkled sees (ggww) from dihybrid X parents: GgWw X GgWw possible gametes GW, Gw, gW, gw, so p (gw)= ¼ one gamete must come from each parents so p (gw*gw)= ¼ * ¼=1/16

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Sum Law: -for outcomes that occur more than one way, mutually exclusive (OR) -ex: what is the chance that you will roll either a 1 or a 6 w/1 die? chance of rolling a 1= 1/6 chance of rolling a 6= 1/6 chance of rolling 1 or 6 = 1/6 + 1/6 = 1/3

E. Chi- Square Analysis 1. Mono & dihybrid ratios are hypothetical predictions based on: -each allele is dom/ recess., segregation is normal, independent asstment. occurs, fertilization is random 2. Chance Deviation: -toss coin for heads/ tails, ratio should be 1:1, but doesn’t usually happen -Null hypothesis: 9:3:3:1, assumes no real difference b/w measured value & predicted; reject/ fail to reject (accept) 2. Chi Square - o(observed)- e (expected) = d (deviation) so x^2 = d/ e (mono) - then find degrees of freedom (df) = n-1 = dif. categories - ex: monohybrid is 3:1 so there are 2 categories, df=1 - analyze table/ chart determine if you can fail to reject - p is a %, if you get .48, 48% of trials would be from chance deviation; must be larger than 5% to be accepted F.

Forked line method 1. Yellow, round (YR) x Yellow round (YR) (F1) use punnett square to determine ratios ¾ are yellow ¾ round ---- 9/16 yellow/ round ¼ wrinkled ---- 3/16 yellow & wrinkled ¼ green ¾ round-------------- 3/16 green and round ¼ wrinkled------------1/16 green & wrinkled G. Polygeneic inheritance 1. Human skin color is a good ex. of polygen. inheritance (multiple genes) 2. assume that 3 dom cap letter genes (A,B,C) ctrl. dark pigment b/c m. melanin is produced 3. a genotype w/all dominant cap genes (AABBCC) has max # of melanin (v. dark skin) 4. geno. w/ all recess. (aabbcc) has lowest amt., v. light skin H. Blood type questions 1. ABO blood groups • each individual is A, B, AB, of O phenotype • phenotype controlled by isoagglutinogen marker on RBC • I^A & I ^B alleles are dominant to the I^O allele • I^A & I^B alleles are codominant to each other Agglutinogens in the ABO Blood system: Blood Type: Type A (AA /AO): Type B (BB/ BO): Type AB (AB): Red blood cell A agglutinogens B aggluts. only A & B aggluts surface proteins only (phenotype) Plasma Antibodies b agglutinin only A agglutinin only No agglutinin (phenotype) Phenotype:

Possible Genotype:

Antigen on RBC surface:

A B AB O

IaIo, IaIa IbIo, IbIb IaIb IoIo

A B AB O

Antibody made in plasma: Anti-B Anti-A Neither Both

Type O (OO): No aggluts a & b aggulitinin

Can donate to:

Can receive from:

A, AB B, AB AB A, B, AB, O

A, O B,O A, B, AB, O O

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Ex: If a male has blood type B and female has type A, what are possible blood types in offspring? • father could be: IbIb, IbIo • mother could be IaIa, IaIo • so child could be: A, B, O Mode of inheritance of Pedigree 1. 2 major categories of genetic diseases • Autosomal & sex linked (almost always X-linked) • 2 subtypes: dominant & recessive 2. Autosomal Recessive • parents generally unaffected • about 25% of offspring are affected if both parents are carriers • 2 affected parents will have an affected child (skips generations) 3. Autosomal Dominant • trait occurs every generation • when one parents it affected, about 50% of offspring will be • affected individs usually heterozygous • unaffected parents don’t produce affected offspring • 2 affected parents can have an unaffected child 4. Sex-linked Recessive (X-linked) • recessive genotype is more common in male • males never pass on their trait to their male offspring • affected female must have affected sons

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affected carrier female (heterozyg.) will have ~ 50% affected male offspring and no affected female offspring, but 50% female offspring will be carriers • ex: colorblindness, more common in males b/c, since male only get one x from mom, if that x is affected w/ color blindness gene, the mall will be color blind Sex Linked Dominant (X-linked) • affected male will always produce affected female offspring and unaffected male offspring (unless mother is affected) • affected female will produce about 50% affected male and 50% affected female if she is heterozygous. If she is homo., 100% of offspring will be affected • no skipping generations Sex Linked (Y chromosome) • trait is always passed father to son, only males affected dominant male (shaded in = recessive, carries trait) circle= female (shaded = recessive) half-shaded circle =carrier but not affected (sex linked) In reading pedigrees: • 1st step: is this ex linked or Autosomal? • if it’s Autosomal: -if individs. is recess. both parents must have at least one recessive allele. Means parents can either be hetero/homozygous -if individ. is dominant, at least 1 parent must have dom. phenol. Can be either homozyg. dom or hetero. • if it’s recessive: -passed on to child from both parents, although parents may seem normal -all kids of 2 affected ppl. are affected, in pedigrees involving rare traits.

J. Linkage questions (coupling, repulsion, how to crunch numbers 1. First, figure out if it’s coupling/ repulsion 2.

Coupling:

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Repulsion:

-most frequent gamete produced will be those w/2 dom, 2 recessives -most frequent gametes are w/ 1 dom. and one recessive

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4. Then to determine the linkage distance divide the number recombinant (those gametes other than those of the parents) gametes into the total games analyzed -ex: parents: pr+ vg+ x pr vg F1: pr+ vg+ (parental, pr+ vg (recombinant), pr vg+ (recomb), pr vg (parental) F1 number: pr+ vg+ : 1339 pr+ vg : 151 pr vg + : 154 pr vg : 1195 total : 2839 gametes 305 (151 pr+vg+ plus pr vg+) gametes = recombinant = 305/2839 x 100= 10.7 cM (centimorgans) = percentage, sort of if less than 50, is linked. Closer to 1, closer together the traits are

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