Genetics Notes week 3
Genetics Notes week 3 Bios 206
Popular in Genetics
verified elite notetaker
Popular in Biological Sciences
verified elite notetaker
This 6 page Class Notes was uploaded by Becca Sehnert on Friday January 29, 2016. The Class Notes belongs to Bios 206 at University of Nebraska Lincoln taught by Dr. Christensen in Fall 2016. Since its upload, it has received 35 views. For similar materials see Genetics in Biological Sciences at University of Nebraska Lincoln.
Reviews for Genetics Notes week 3
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 01/29/16
Week 3 notes LECTURE Monday Epistasis –when genotype still follows Mendel’s laws, but phenotypes are more complex • When one allele of gene masks effect of another gene • 2 gene pairs interact where 1 allele is required at each locus to express a certain phenotype o Locus –location, map trait to location on chromosome o Locus = gene • Still used forked-line method or Punnett square as long as they assort independently • Agouti o C gene codes for color o With out C gene (cc), albino mouse o Without A gene (aa), black pigment always on A_C_ = agouti, aaC_ =black, --cc = albino A female aacc mouse was impregnated by an unknown male. Litter produced 5 agouti and 5 albino mice. What was genotype of male? a. AAcc b. AaCC c. AACc d. AaCc e. aaCc • Cats! o wwA_ o W_ _ _ No cells that can make pigment. Often deaf o wwaa Simple recessive mutation. Knock out agouti gene Fruit chape in summer squash are determined by 2 known genes, A and B. A_B_ squash are disc, aaB_ and A_bb are spehere shapped, aabb are long. 2 sphere shaped squash crossed and progeny are all disc. What were geno of parents? a. AAbb and aaBB b. AABB and aabb c. AaBb and AaBb Phenotypes can be affected by more than one gene • Many traits with distinct phenotype are affected by more than 1 gene • In gene interaction, function of gene products contribute to development of phenotype • Bombay phenotype o Example of epistasis where homo recessive allele of 1 gene masks expression of second gene o Behave like type O cuz don’t make h substance o Without h substance, are type O o PRACTIC GENOTYPING Examples of epistasis modifying dihybrid ratios • If dihybrid cross, remember 9331 ratio • If test cross, understand how genes interact and produce pheno X-linkage describes genes on X-chromosome • Genes on x chrom have unique patterns of inheritance • Due to presence of one x chrom in males • Males are hemizygous for x-linked genes • A_oo Agouti gene, see pattern. • O stands for orange on x chrom. If not orange, oo is geno • ?? OY Happens in spite of agouti. Follows tabby gene rules • A_Oo • In drosophila, double expression of x gene to mimic females • Female mammals mosaics X = aaOo • Drosophila eye color o White eyed mutants –don’t see well o White is x-linked o See phenotype of female if fly is male Vermillion eye color mutation (v-) is x-linked and recessive in Drosophila Cross true-breeding vermilion femals to true-breeding wild type males will produce what ratio in F1? WTf:vf:WTm:vm a. 1:1:1:1 b. 1:0:0:1 c. 0:1:0:1 d. 1:1:0:0 e. 1:1:1:0 Cross F1 vermilion males to wild females in previous question will product… WTf:vf:WTm:vm a. 1:1:1:1 b. 1:0:1:0 c. 0:1:0:1 d. 1100 e. 1110 LECTURE WEDNESDAY New gene or new allele??? • Do a complementation test • Works for recessive mutation • Complementation analysis is done by mating 2 homo single mutant strains and examining the F1 progeny o Have to know recessive and homozygous o Have to know that it’s a single mutant strain (find this by crossing with wild-type) CCpp ccPP =CcPp CP Cp cP cp CP CCPP CCPp CcPP CcPp Cp CCPp CCpp CcPp Ccpp cP CcPP CcPp ccPP ccPp cp CcPp Ccpp ccPp ccpp COMPLEMENTATION Flower color in sweet peas is determined by at least 2 genes, C and P. If either gene is homo for recessive allele (cc or pp), flowers are white. A new white wlowered mutation is discovered. Tests determine that it is due to single recessive mutation, and true- breeding line established. Crossing new white mutant to ccPP produces progeny who ALL have white flowers. You can conclude that the new mutant: a. Is allele of gene C b. Is not an allele of gene C c. Is an allele of gene P d. Is not an allele of gene P e. Is an allele of neither genes C nor P Eye pigments in Drosophila • 2 pigments in wild type eye, brown and red. Combination makes brick-red eye • Mutations in 1 gene required for synthesis of brown pigment leave bright red • Mutation in gene for synthesis of red pigment leave brown eyes • Double mutants have neither pigment, thus white eyes • Pathways…many genes, one phenotype o 2 mutants, both affecting same pathway, same phenotype Complemenation test Unk/unk x aa à bright red Failure to complement- unk must be allele of gene A Unk/unk x zz à wild type Complementation! Unk must NOT be allele of gene A, something else like gene B Flower color in sweet peas is determined by at least 2 genes, C and P. If either gene is homo for recessive allele (cc or pp), flowers are white. A new white wlowered mutation is discovered. Tests determine that it is due to single recessive mutation, and true- breeding line established. Crossing new white mutatnt to CCpp produces progeny who ALL have purple flowers. You can conclude that new mutant: a. Is an allele of gene C b. Is not an allele of gene P c. Is an allele of gene P d. Is not an allele of gene P e. Is an allele of neither genes C or P Both cross parents are white, so has to have another allele affecting the phenotype like aa or something. So unknown would be something like PPaa since ALL are purple (cross inhibiting expression of pp and new mutation, aa) Flower color in sweet peas is determined by at least 2 genes, C and P. If either gene is homo for recessive allele (cc or pp), flowers are white. A new white wlowered mutation is discovered. Tests determine that it is due to single recessive mutation, and true- breeding line established. Crossing the new white mutant to ccpp produces progeny who ALL have purple flowers. Can conclude that new mutant: a. Is an allele of gene C b. Is not an allele of gene P c. Is an allele of gene P d. Is not an allele of gene P e. Is an allele of neither genes C or P Alleles of same gene aa x bb = ab (all mutant) F2 from self cross ¼ aa, ½ ab, ¼ bb (all mutant) FRIDAY LECTURE Pleiotropy • Single gene has multiple phenotypes • Osteogenesis imperfect: brittle bones AND blue sclera • B-globin mutation:; sick cell disease AND malaria resistance. Recessive Phenotype may be influenced by environment and genotype Variable expressivity –all individuals of same genotype show phenotype but to varying degrees (50 broken bones vs. 7) Incomplete penetrance –not all individuals with same genotype display phenotype. NOT A CATCH ALL FOR EXAMS Ss Ss SS SS C_aa C C aa –warmer parts cannot make pigment c c aa c c aa Temperature sensitive mutations Conditional/temp-sensitive Nutritional mutations Found in model organisms In biosynthetic in bacteria or yeast Can be conditional lethal –cell will die without added nutrient Age of onset and imprinting are examples of how phenotype Not much on pedigrees
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'