AB AB AB AB
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BIOL 2416 GENETICS Dr. A. Tineke BerendsTWO-POINT demo CROSSES WHY?1. 2.WHAT? AUTOSOMAL RECESSIVE double heterozygote X homozygous recessive dominant = + symbolized by a +and b = wild-typea+b+/ab x ab/abTo determine if 2 genes are linked lớn estimate the bản đồ distance between 2 linked genes (more accurate if genes are close together và when large numbers of progeny are scored)AUTOSOMAL DOMINANT double heterozygote X homozygous recessive mutant alleles are dominant, so recessive alleles = + wild type = A andB+ AB/A+B+ x A+B+/A+B+X-LINKED RECESSIVE female double heterozygote X male hemizygous for wild-type alleles + dominant = a andb+ = wild-typeX-LINKED DOMINANT female double heterozygote X Male hemizygous for wild-type alleles recessive = wild+ + type = A & Ba+b+/ab x ab/YAB/A+B+ x A+B+/YNOTE: The baby phenotypes will depend on how the linked alleles are arranged in the heterozygous parent! In the above examples, both dominant alleles are arranged on one chromosome, while the recessive alleles are arranged on the other chromosome (called CIS coupling). But you could also put one dominant and one recessive on each chromosome (e.g. A+b/ab+; called TRANS coupling). HOW? e.g. For the 2-PT thử nghiệm cross (using cis-coupling of an autosomal recessive trait): phường genotypes: p. Phenotypes:a+b+/ab a+b+xab/ab ab1Note we are using the allele symbols lớn represent PHENOtypes. From this cross, you might get the following progeny phenotypes (baby flies): Baby phenotype + +ab ab a+b ab+Looks like? Look lượt thích 1st parentObserved # of babies 965Look like 2nd parent Recombinant type944 206Recombinant type185 Total = 2300 babies1.HOW CAN YOU TELL FROM THESE NUMBERS THAT THE A and B gene ARE LINKED ON THE SAME CHROMOSOME?IF the genes were NOT linked (carried on different chromosomes), the heterozygous parent could size the following gametes in EQUAL proportions (allowed to lớn FOIL lớn find gametes): + + + + 1 a b : 1 ab : 1 a b : 1 ab That means you would EXPECT lớn see the following numbers (based on the same total # of actual babies) in the offspring: Baby phenotypeLooks like?Observed # of babiesa+b+ ab a+b ab+Look like 1st parent965Expected # of babies IF genes are NOT linked 2300/4 = 575Look like 2nd parent Recombinant type944 2062300/4 = 575 2300/4 = 575Recombinant type1852300/4 = 575Total = 2300 babiesTotal = 2300 babiesThe fact that the actual number pattern does NOT match the expected number pattern for unlinked genes, means the genes are NOT carried on different chromosomes.
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This implies we are dealing with LINKED genes. But does the actual number pattern match the case where we have two genes that stay linked all the time? If that were the case, the heterozygous (cis) parent would only be able lớn make the following gametes, since NO FOILING is allowed: + +1 a b : 1 ab2Then we would expect to see the following: Baby phenotypeLooks like?Observed # of babiesa+b+ ab a+b ab+Look like 1st parent965Expected # of babies IF genes are linked ALL THE TIME (no crossover at all) 2300/2 = 1150Look lượt thích 2 parent Recombinant type944 2062300/0 = 1150 0Recombinant type1850Total = 2300 babiesTotal = 2300 babiesndObviously, the actual number pattern does not match the “totally linked” scenario either. We seem khổng lồ have a pattern that falls somewhere in between the UNLINKED & TOTALLY LINKED scenarios. The only way to explain this is to assume the two genes ARE LINKED, but are sometimes “unlinked” by a single crossover event between them.2. NOW THAT WE KNOW THE TWO GENES ARE LINKED, HOW FAR APART ARE THEY ON THE SAME CHROMOSOME? Assuming that the further the 2 genes are apart, the greater the chance that crossover will occur between them, we can use the (single) crossover frequency as an estimate of bản đồ distance between the genes. Since the only way you can get a recombinant type baby is by using crossover gametes in the heterozygous parent (the ones you can only get by foiling), we can use the numbers (%) of recombinant babies khổng lồ estimate map distances: (# of observed recombinant babies / total # of babies)(100%) = % crossover events = approximate map distance between the 2 genes (in bản đồ units or cM) So for the above example: (206 + 185)/2300 x 100% = 17 % crossover = genesa+ and b+ are 17 maps units or cm apart 32-POINT demo CROSS METHOD RECAP: 1.Determine the genotypes of the demo cross parents (this will depend on the inheritance pattern of the trait; we will mostly use autosomal recessive; one parent is heterozygous, the other is homozygous recessive).2.Use allele symbols lớn represent the phenotypes of the test cross parents (the heterozygous parent will be dominant-dominant, the homozygous recessive parent will be recessive-recessive).3.Look at the danh sách of observed phenotypes in the babies. What kind of pattern bởi vì you see?OBSERVED PATTERN 1:1:1:1 1:1 More parental types than recombinant types.CONCLUSION The two genes are NOT linked. The two genes are linked and STAY linked ALL of the time. The two genes are linked, but there is some crossover between them.4.If desired, vị a Chi-square test, comparing the observed pattern with the 1:1:1:1 pattern (you will have to calculate expected progeny numbers using the same total number of baby flies). You will either “reject” or “fail lớn reject” the hypothesized 1:1:1:1 pattern.5.If you have linked genes with crossover: (# observed recombinants/# observed total progeny)/100% = % crossover = map distance in map units or cm between the genes.4
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BIOL 2416 GENETICS Dr. A. Tineke BerendsTWO-POINT demo CROSSES WHY?1. 2.WHAT? AUTOSOMAL RECESSIVE double heterozygote X homozygous recessive dominant = + symbolized by a +and b = wild-typea+b+/ab x ab/abTo determine if 2 genes are linked lớn estimate the bản đồ distance between 2 linked genes (more accurate if genes are close together và when large numbers of progeny are scored)AUTOSOMAL DOMINANT double heterozygote X homozygous recessive mutant alleles are dominant, so recessive alleles = + wild type = A andB+ AB/A+B+ x A+B+/A+B+X-LINKED RECESSIVE female double heterozygote X male hemizygous for wild-type alleles + dominant = a andb+ = wild-typeX-LINKED DOMINANT female double heterozygote X Male hemizygous for wild-type alleles recessive = wild+ + type = A & Ba+b+/ab x ab/YAB/A+B+ x A+B+/YNOTE: The baby phenotypes will depend on how the linked alleles are arranged in the heterozygous parent! In the above examples, both dominant alleles are arranged on one chromosome, while the recessive alleles are arranged on the other chromosome (called CIS coupling). But you could also put one dominant and one recessive on each chromosome (e.g. A+b/ab+; called TRANS coupling). HOW? e.g. For the 2-PT thử nghiệm cross (using cis-coupling of an autosomal recessive trait): phường genotypes: p. Phenotypes:a+b+/ab a+b+xab/ab ab1Note we are using the allele symbols lớn represent PHENOtypes. From this cross, you might get the following progeny phenotypes (baby flies): Baby phenotype + +ab ab a+b ab+Looks like? Look lượt thích 1st parentObserved # of babies 965Look like 2nd parent Recombinant type944 206Recombinant type185 Total = 2300 babies1.HOW CAN YOU TELL FROM THESE NUMBERS THAT THE A and B gene ARE LINKED ON THE SAME CHROMOSOME?IF the genes were NOT linked (carried on different chromosomes), the heterozygous parent could size the following gametes in EQUAL proportions (allowed to lớn FOIL lớn find gametes): + + + + 1 a b : 1 ab : 1 a b : 1 ab That means you would EXPECT lớn see the following numbers (based on the same total # of actual babies) in the offspring: Baby phenotypeLooks like?Observed # of babiesa+b+ ab a+b ab+Look like 1st parent965Expected # of babies IF genes are NOT linked 2300/4 = 575Look like 2nd parent Recombinant type944 2062300/4 = 575 2300/4 = 575Recombinant type1852300/4 = 575Total = 2300 babiesTotal = 2300 babiesThe fact that the actual number pattern does NOT match the expected number pattern for unlinked genes, means the genes are NOT carried on different chromosomes.
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This implies we are dealing with LINKED genes. But does the actual number pattern match the case where we have two genes that stay linked all the time? If that were the case, the heterozygous (cis) parent would only be able lớn make the following gametes, since NO FOILING is allowed: + +1 a b : 1 ab2Then we would expect to see the following: Baby phenotypeLooks like?Observed # of babiesa+b+ ab a+b ab+Look like 1st parent965Expected # of babies IF genes are linked ALL THE TIME (no crossover at all) 2300/2 = 1150Look lượt thích 2 parent Recombinant type944 2062300/0 = 1150 0Recombinant type1850Total = 2300 babiesTotal = 2300 babiesndObviously, the actual number pattern does not match the “totally linked” scenario either. We seem khổng lồ have a pattern that falls somewhere in between the UNLINKED & TOTALLY LINKED scenarios. The only way to explain this is to assume the two genes ARE LINKED, but are sometimes “unlinked” by a single crossover event between them.2. NOW THAT WE KNOW THE TWO GENES ARE LINKED, HOW FAR APART ARE THEY ON THE SAME CHROMOSOME? Assuming that the further the 2 genes are apart, the greater the chance that crossover will occur between them, we can use the (single) crossover frequency as an estimate of bản đồ distance between the genes. Since the only way you can get a recombinant type baby is by using crossover gametes in the heterozygous parent (the ones you can only get by foiling), we can use the numbers (%) of recombinant babies khổng lồ estimate map distances: (# of observed recombinant babies / total # of babies)(100%) = % crossover events = approximate map distance between the 2 genes (in bản đồ units or cM) So for the above example: (206 + 185)/2300 x 100% = 17 % crossover = genesa+ and b+ are 17 maps units or cm apart 32-POINT demo CROSS METHOD RECAP: 1.Determine the genotypes of the demo cross parents (this will depend on the inheritance pattern of the trait; we will mostly use autosomal recessive; one parent is heterozygous, the other is homozygous recessive).2.Use allele symbols lớn represent the phenotypes of the test cross parents (the heterozygous parent will be dominant-dominant, the homozygous recessive parent will be recessive-recessive).3.Look at the danh sách of observed phenotypes in the babies. What kind of pattern bởi vì you see?OBSERVED PATTERN 1:1:1:1 1:1 More parental types than recombinant types.CONCLUSION The two genes are NOT linked. The two genes are linked and STAY linked ALL of the time. The two genes are linked, but there is some crossover between them.4.If desired, vị a Chi-square test, comparing the observed pattern with the 1:1:1:1 pattern (you will have to calculate expected progeny numbers using the same total number of baby flies). You will either “reject” or “fail lớn reject” the hypothesized 1:1:1:1 pattern.5.If you have linked genes with crossover: (# observed recombinants/# observed total progeny)/100% = % crossover = map distance in map units or cm between the genes.4
