Lecture 17: Mendelian Genetics PDF

Lecture 17 Mendelian Genetics Required Reading Chapter 14 - Section 14.6 Human Genetics (pp. 313 to 315) Objectives: Monohybrid vs. Dihybrid Crosses Test Crosses Incomplete Dominance and Epistasis Textbook reference sections/pages: Morris text – Chapter 14 (pp...

Lecture 17 Mendelian Genetics Required Reading Chapter 14 - Section 14.6 Human Genetics (pp. 313 to 315) Objectives: Monohybrid vs. Dihybrid Crosses Test Crosses Incomplete Dominance and Epistasis Textbook reference sections/pages: Morris text – Chapter 14 (pp. 303-315) Section 14.4 Segregation(pp. 303-307) Section 14.5 Independent Assortment (pp. 309 – 313) BIOL1010 and 1  A quick review…. How many genes are shown on this chromosome? BIOL1010 and 2  Genetics Terminology P1 generation – parental generation F1 generation – first offspring from the parental generation – F2, F3 etc. Alleles – different forms of a gene Homozygous - having two of the same allele for a given gene Heterozygous – having two different alleles for a given gene Genotype – the combination of alleles in an individual Phenotype – the expression of a trait in an individual Dominant – the trait that is phenotypically expressed in a heterozygous individual Recessive – the trait that is phenotypically expressed when two copies of the same allele are present 3 BIOL1010 and  True-Breeding One parent will only make gametes with the A allele, whereas the other parent will make gametes with the a allele. All of the offspring (F1 generation) will be heterozygous, Aa, for the gene and show the dominant phenotype. BIOL1010 and 4 Segregation: The Monohybrid Cross True-breeding parents (P1 generation) All of the offspring (F1 generation) will be heterozygous, Aa, for the gene and show the dominant phenotype. Cross of F1 progeny monohybrids i.e. heterozygous for one character Resulting phenotypic ratio is 3:1 Resulting genotypic ratio is 1:2:1 Genotypic ratios can be BIOL1010 and different from phenotypic 5  Segregation BIOL1010 and 6  Practice Problem In certain plants, tall is dominant to short. If a heterozygous plant is crossed with a homozygous tall plant, what is the probability that the offspring will be short? A) 1 B) 1/2 C) 1/4 D) 1/6 E) 0 BIOL1010 and 7  Practice Problem In humans, being a tongue roller (R) is dominant over non- roller (r). A man who is a non-roller marries a woman who is heterozygous for tongue rolling. What is the probability of this couple having a child who is a tongue roller? BIOL1010 and 8  The Testcross What if you were given a yellow seed and asked to determine the genotype? Yellow is dominant, so genotype can be AA or Aa Perform a testcross: – Cross the yellow seed with a green seed and observe the phenotype of the seeds resulting from the cross Why use a green seed in the cross? BIOL1010 and 9  Practice Problem A pet angelfish breeder is looking to breed her champion black lace angelfish. She knows that black colour is dominant to white but doesn’t know if her angelfish is purebred. How can she find out? 10 BIOL1010 and  Degrees of Dominance Mendel’s classic pea experiments: – The F1 offspring always looked like one of the two parental varieties – Due to one allele showing complete dominance over the other – But, many genes do not follow Mendelian inheritance Degrees of Dominance – Alleles can show different degrees of dominance and recessiveness with relation to each other BIOL1010 and 11  Degrees of Dominance Incomplete dominance – The phenotype of F1 hybrids is somewhere between the phenotypes of the two parental varieties – Neither red or white allele is completely dominant – Flowers of the heterozygous have less red pigment than red homozygous dominants The original traits show up again when you cross the F1’s BIOL1010 and 12  Practice Problem An Angora rabbit (long fur) is crossed with a short- furred Rex rabbit. All of the offspring have medium length fur. What are the genotypes of the parent rabbits? What is the mode of inheritance? What could you do to confirm this? 13 BIOL1010 and  Independent Assortment The principle of independent assortment states that segregation of one set of alleles of a gene pair is independent of the segregation of another set of alleles of a different gene pair. The hereditary transmission of either gene has no effect on the hereditary transmission of the other. 14 BIOL1010 and  Independent Assortment Illustrated 15 BIOL1010 and  Dihybrid Cross Demonstrates Independent Assortment Allele for yellow seeds (Y) is dominant to allele for green RrYy seeds (y). Allele for round seeds (R) is dominant to allele for wrinkled seeds (r). Resulting Phenotypic RrYy Ratio is 9:3:3:1 BIOL1010 and 16  Determining Gametes What gametes are possible for each parent in the following crosses? (assume independent assortment): AaBB x Aabb AABBCC x aabbcc BIOL1010 and 17  Determining Probability of Offspring For the following crosses, determine the probability of obtaining the indicated genotype Crossin an offspring Offspring Probability AAbb x AaBb AAbb AaBB x AaBb aaBB AABbcc x AaBbCc aabbCC BIOL1010 and 18 Parents Offspring AAbb x AaBb AAbb 19 BIOL1010 and Parents Offspring Parents Offspring AAbb x AaBb AAbb AaBB x AaBb aaBB Parents Offspring AABbcc x aabbCC AaBbCc 20 BIOL1010 and  Determining Probability of Offspring For the following crosses, determine the probability of obtaining the indicated genotype in an offspring Cross Offspring Probability 1/4 AAbb x AaBb AAbb 1/8 AaBB x AaBb aaBB AABbcc x 1/2 AaBbCc aabbCC BIOL1010 and 21  Practice Problem In guinea pigs, the brown coat colour allele (B) is dominant over red (b) and the solid colour allele (S) is dominant over spotted (s). The F1 offspring of a cross between true- breeding brown, solid-coloured guinea pigs and red, spotted guinea pigs are crossed. What proportion of their offspring (F2) would be expected to be red and solid coloured? A. 1/9 B. 1/16 C. 3/16 D. 9/16 E. 3/4 BIOL1010 and 22  Practice Problem In guinea pigs, the brown coat colour allele (B) is dominant over red (b) and the solid colour allele (S) is dominant over spotted (s). The F1 offspring of a cross between true-breeding brown, solid- coloured guinea pigs and red, spotted guinea pigs are crossed. What proportion of their offspring (F2) would be expected to be red and solid coloured? A. 1/9 B. 1/16 BS Bs bS bs C. 3/16 BS BB BbS Bb BB D. 9/16 SS Ss S Ss E. 3/4 Bs BB BBs Bb Bbs Ss s Ss s P1: BBSS P2: bbss bS Bb Bb bbS bbS F1: BbSs Ss Ss S s F2:? BbSs x bs Bb Bbs bbS bbs BbSs Ss s s s 23 BIOL1010 and  Genes That Modify Phenotype Epistasis – Interaction of genes can affect the same trait – How? The products of two or more genes result in a certain phenotype; or The product of one gene masks or changes the expected phenotype of one or more other genes See modification of expected ratio! – Textbook example: Chickens (Figure 14.25 p.312) Must know this example! – Colouration in Labrador Retrievers is an example of epistasis BIOL1010 and 24  Epistasis in Labrador Retrievers B= black fur BbEe BbEe b=brown fur Sperm E= with pigment 1 /4 BE 1 /4 bE /4 1 Be 1 /4 be e= without pigment Eggs 1 /4 BE BBEE BbEE BBEe BbEe E is epistatic gene 1 /4 bE BbEE bbEE BbEe bbEe One gene determines color and one gene (E, /4 1 Be BBEe BBee Bbee epistatic gene) BbEe determines if there is /4 1 be pigmentation or not BbEe bbEe Bbee bbee Note: Labrador with 9 : 3 : 4 “ee” results in having no pigmentation! BIOL1010 and 25 Epistasis in Labrador Retrievers Epistasis in Labrador Retrievers – In this example, a gene at one locus alters the phenotypic expression of a gene at a second locus. – One gene determines whether the hair pigment is black (B) or brown (b). The black allele (B) is dominant to the brown allele (b). – Another gene, the epistatic gene, determines whether the dogs will have pigment deposited in hair (E) or not (e) The pigment allele (E) is dominant to absence of (e) pigment. – An (ee) individual has no pigment (i.e. yellow) regardless of the genotype of the first gene. BBee, Bbee, bbee all yellow labs BIOL1010 and 26  Practice Problem Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether or not cactuses have spines. Homozygous recessive nn cactuses have no spines at all. If doubly heterozygous SsNn cactuses were allowed to self- pollinate, the offspring would segregate in which of the following ratios? A) 3 sharp-spined:1 spineless B) 1 sharp-spined:2 dull-spined:1 spineless C) 1 sharp-spined:1 dull-spined:1 spineless D) 1 sharp-spined:1 dull-spined E) 9 sharp-spined:3 dull-spined:4 spineless BIOL1010 and 27  Practice Problem F1: SsNn X SsNn SN Sn sN sn F2 :? SN SSN SS SsN SsN Recall : nn have N Nn N n no spines Sn SS SSn SsN Ssn Nn n n n sN SsNN SsN ssN ssN n N n sn SsN Ssn ssN ssn n n n n F2 phenotypically would show: 9 sharp spines 3 dull spines 4 no spines EPISTASIS ! 28 ( answer is E) BIOL1010 and