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Lecture 9

BIOC15Fall2013 Lecture 9 and Lecture 10 Notes.docx

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Department
Biological Sciences
Course
BIOC15H3
Professor
Karen Williams
Semester
Fall

Description
BIOC15Fall2013 Lecture 9 Notes: Sex linkage maps and inheritance and Lecture 10: Linkage and chromosomes How do incomplete domninance and codominance affect phenotypic ratios? o ABO blood group  in order for the child to have O/O blood, each parent must have at least one O o o having AB blood group is an example of co-dominance because your body makes both the A and B antigens Multiple alleles – white locus in Drosophila o there are over 100 alleles of white locuts (w)  recessive o if white eosin is an allele of white then: e o white eosin (w ) should be on the same chromosome as white (w) o w females crossed with w males should give the mutant phenotype o o both the female and male F1 have eosin eyes o The F1 female produced, is then crossed with red-eyed (wt) male: o o the males have white eyes, and the females do not because the males only have one X chromosome, so complementation of genes does not occur o but F1 progeny all have eosin eyes  this is due to the fact that mutation occurs in the same gene for both parents, so their genes cannot come together to form a wt offspring o o an example of complementation  crossing a black fly (b+/b+ e/e) with a black fly (b/b e+/e+) will give rise to a wt coloured fly because the mutations, causing the same phenotypes occur in different genes and so will result in a heterozygous offspring, which will have wt phenotype Seed coat pattern in lentils are determined by a gene with 5 alleles o spotted (C ), dotted, clear, marbled-1 (C ) and marbled-2 (C ) 1 o o F2 Mendelian ratio is 1:2:1 o the ratio are determined by alternative alleles of a single gene  heterozygotes have phenotype different than either homozygote o so reciprocal crosses between pairs of pure breeding lines is used to determine dominance relations o marbled 1 > marbled 2 > spotted = dotted > clear o so spotted and dotted are codominant Novel phenotypes resulting from gene interactions – Seed coat in lentils o dyhibrid cross of lentils, tan x gray o all F1 seeds are brown o F2 progeny = 9/16 brown, 3/16 tan, 3/16 gray and 1/16 green o 9:3:3:1 phenotypic ratio  suggests independently assorting genes for seed coat colour o o this 1:1:2:2:1:1:2:2:4 F2 genotypic ratio corresponds to the 9:3:3:1 F2 phenotypic ratio Sorting out the dominance relations by select crosses of lentils o F2 phenotypes from dihybrid crosses ill be 9:3:3:1 ratio only when dominance of alleles of both genes is complete o Complementary gene action generates purple flower colour in sweet peas o dihybrid cross generates a 9:7 ratio in F2 progeny  9 purple and 7 white 2 o o only one locus is needed to be homozygous recessive for white colour to appear  A_bb, aaB_, aabb) o possible biochemical explanation for complementary gene action in sweet peas: o o one pathway has 2 reactions catalyzed by 2 different enzymes o at least one dominant allele of both genes is required for purple pigment, as this ensures both enzymes are present and active o homozygous recessive for either or both genes results in no pigment, so white flowers Recessive epistasis in Golden Labrador dogs o 9:3:4 ratio in F2 progeny of dihybrid crosses indicates recessive epistasis o 9 black (B_ E_) o 3 are brown (bb E_) o and 4 are yellow (B_ ee, bbee)  genotype ee masks the effect of all B genotypes (this is called recessive epistasis: where being homozygous recessive for one gene, masks the effect of another gene) Modified 9:3:3:1 ratio – Labradors o B/_ black labs, bb chocolate brown lab o _/_ ee = golden yellow lab o black lab x brown lab  BBee x bbEE o F1 were all black labs  BbEe o F2 gave 9:3:4 ratio of 9 black labs, 3 brown labs and 4 golden labs  B_E_ = black (9), bbE_(3) = brown and B_ee(3) and bbee(1) were golden (so 4 golden in total) o Epistasis  2 or more genes interact to control a phenotype Coat colour in mice 0 determined by multiple alleles of several genes o The A gene: A= agouti, a= nonagouti o The B gene: B= black pigment, b= brown o The C gene: C= pigmented, c= albino o The D gene: D=nondilution, d= dilution o The S gene: S= unspotted, s= pigmented spots on a white background o Agouti pattern: yellow band around each of the hairs, wildtype pattern o AA x aa  Aa x Aa  1 AA (agouti): 2 Aa (agouti): 1aa (nonagouti)  3:1 agouti : non agouti o A is dominant, but it is recessive if homozygous to other genes because AA is lethal o Black pigment hair is the wildtype hair colour and B is dominant to b which confers brown hair colour o BB x bb  1 BB : 2 Bb : 1 bb  3:1 black to brown o Albino mice are homozygous for the recessive member of the C/c allelic pair  the c/c constitution is epistatic to the other coat colour genes o So unless c/c, mice can have any colour and combination of colour coats, but when individual is cc at the C allele, it will be white o These 3 genes combined can give rise to many different combinations of coat colour, and mice have about 25000 genes in total Genetic heterogeneity in humans: mutations in many genes can cause deafness 3 o two deaf parents can have hearing offspring due to genetic complementation (homozygous when parents carry mutations in the same gene, complementation does not occur and thus the child will be deaf Two hypotheses to
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