Answer part a)According to the gene map of the wild type and the m.pdf

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Answer part a) According to the gene map of the wild type and the mutant gene, it can be seen that the mutant lacks a site for restriction digestion. This shows that in wild type, the restriction enzyme can digest the DNA at any three sites and thus atleast two bands will appear after amplification. On the contrary, the deletion of middle sequence of restriction site will lead to formation of a single band in mutant gene. The smaller band will represnt the mutant and the larger band will represent the normal/wild type allele. Thus, according to this information, it can be deciphered that the mother is a heterozygous in nature and carries a mutant allele for Tay-Sachs disease while the father does not. Thus, the lower band will be a representative of mutant or \"t\" and the upper band will be a representative of normal \"T\" allele. Answer part b) Since the mother is a carrier, there are 50% chances that the mother will transmit the mutant allele to her offspring. Thus, the couple can bear a normal heterozygous carrier female offspring or an affected male offspring. According to the information in the image, the genotype bands of father and offspring match which clearly suggest that the offspring is an unaffected heterozygous carrier female. Answer part c) There are always 50% changes for a male to be affected by the disease. Rest 50% males will be normal both geotypically and phenotypically. Also, all the females will be phenotypically normal but 50% of them would be heterozygous carriers. Genotypes: XtX : XtY : XX : XY :: 1:1:1:1 Phenotypes: Affected : normal : carriers :: 1: 2: 1XYXtXtXXtYXXXXY Solution Answer part a) According to the gene map of the wild type and the mutant gene, it can be seen that the mutant lacks a site for restriction digestion. This shows that in wild type, the restriction enzyme can digest the DNA at any three sites and thus atleast two bands will appear after amplification. On the contrary, the deletion of middle sequence of restriction site will lead to formation of a single band in mutant gene. The smaller band will represnt the mutant and the larger band will represent the normal/wild type allele. Thus, according to this information, it can be deciphered that the mother is a heterozygous in nature and carries a mutant allele for Tay-Sachs disease while the father does not. Thus, the lower band will be a representative of mutant or \"t\" and the upper band will be a representative of normal \"T\" allele. Answer part b) Since the mother is a carrier, there are 50% chances that the mother will transmit the mutant allele to her offspring. Thus, the couple can bear a normal heterozygous carrier female offspring or an affected male offspring. According to the information in the image, the genotype bands of father and offspring match which clearly suggest that the offspring is an unaffected heterozygous carrier female. Answer part c) There are always 50% changes for a male to be affected by the disease. Rest 50% males w.

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Answer part b)
Since the mother is a carrier, there are 50% chances that the mother will transmit the mutant
allele to her offspring. Thus, the couple can bear a normal heterozygous carrier female offspring
or an affected male offspring. According to the information in the image, the genotype bands of
father and offspring match which clearly suggest that the offspring is an unaffected heterozygous
carrier female.
Answer part c)
There are always 50% changes for a male to be affected by the disease. Rest 50% males will be
normal both geotypically and phenotypically. Also, all the females will be phenotypically normal
but 50% of them would be heterozygous carriers.
Genotypes: XtX : XtY : XX : XY :: 1:1:1:1
Phenotypes: Affected : normal : carriers :: 1: 2: 1XYXtXtXXtYXXXXY

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a) How are X-linked genes inherited in humans? Be able to use this information to perform monohybrid and dihybrid crosses and to interpret pedigrees. (SHOW WORK & EXPLAIN) b) Describe dosage compensation/X-inactivation and its consequences. (SHOW WORK & EXPLAIN) c) Describe Y-linked inheritance and be able to use this information in monohybrid and dihybrid crosses and pedigrees. (SHOW WORK & EXPLAIN) d) Explain complete dominance, incomplete dominance, codominance, penetrance, expressivity, lethal alleles and multiple alleles and give examples of each. (SHOW WORK & EXPLAIN) Solution If cross is made between XY MALE and Xx carrier female then offspring will as shown below 50% hemophilic Xx ( carrier) female and hemophilic boy Yx 50% normal XX and XY. 2.In many animals male have one and female have two X chromosome . The difference in X chromosome dosage between the two sexes is compensated by mechanism that regulate X chromosome transcription. Mainteneance of chromosome dosage is necessary in normal development. X chromosome contain many gene that are important in both sexes male contain single X and female contain two X chromosome. .To compensate for X chromosome dosage difference between sexes different mechanism have evolved to equalize x linked transcript level in male and female.. polypoidy refers to increase in dosage in all chromosome while aneuploidy refers to change in dosage of one chromosome with respect to rest of the genome. Consequence of X gene inactivation is visibly mosaic phenotypes exhibited by mammalian female heterozygous for X linked genes affecting hair and skin . Answer 3: Y linked inheritance can be seen in male members as only male has Y chromosome. Y-linked inheritance is appearance of traits in an organism due to the presence of an alelle on the Y-chromosome e.g hypertrichosis and pattern baldness are due to Y linked inheritance. Answer 4; Complete dominance is form of dominance in hetrozygous allele that is dominant completely mask the effect of allele that is recessive flower on mendel\'s pea plant is example of complete dominance Blue flower has dominant allele B whle b is recessive cross will be dominant are BB, Bb,Bb with blue flowers Incomplete dominance is a form of inheritance in which one allele for a specific trait is not completely expressed over its paired allele .As a result third phenotype appear which is the combination of both allele e. g snapdragon flower is pink when cross between red flower and white flower here neither white is dominant nor red Codominance is form of dominance where in allele of a gene pair in a heterozygote are fully expressedXYXXXXYxXxxY.

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Can yyou answer number 5 please advance genetic As discussed in class, the dark (melanic) form of the peppered moth, Biston betularia, has a survival advantage in industrialized regions. The melanic allele is dominant to the typical (light) allele. In a particular forest within an industrialized region, the frequency of the melanic allele is 0.7 and the typical allele is 0.3. The light form of moths have a reproductive success that 47% that of the dark form. What will the allele frequencies be after one generation of selection? Given a population that up to now had been in Hardy-Weinberg equilibrium. Assume two alleles, one locus, rho = 0.5, and distinctly different (and unambiguous) phenotypes associated with each genotype. Now assume internal fertilization and that all matings over one generation are 100% assortative with regard to the trait in question. What are the genotype frequencies before the round of assortative mating? What are the genotype frequencies in the generation that follows this round of assortative mating? Solution A) The genotype frequencies during the Hardy-Weinberg equlibrium must be favouring heterozygosity i.e. any individual with any given genotype has the equal chance of mating with another individual of any given genotype. The equlibrium of individual allele frequencies and the proportion of the various genotypic combinations is established when p value statistically significant value is 0.5 the disimilar genotypic populations mates together. B) Assortive mating leads to an over abundance of homozygous individuals Which are likely to share similar genotypes as When matings of similar phenotypes occur more frequently than by random chance, the likelihood of offspring receiving two copies of an identical allele increases, disrupting the Hardy-Weinberg expectations. There will be the increase in total population variance as homozygosity will be increased for example if we have a trait controlled by 2 loci then at equilibrium would deviate one from another significantly. Assortative mating can act as a powerful mechanism for genetic change in a population specially where phenotype can be attributed to the interaction at 2 loci with no dominance, it is easier that 100% assortative mating results in an overall increase in homozygosity and total population variance thus affecting genotype frequencies..

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a. From the DNA fragmentation analysis, it is clear that the polydactyly came from the parents X chromosome as the affected sons have same band pattern as father has one band that found in sons and mother has two bands that appeared in affected sons. The father genotype is XbY, where Xb is affected with polydactyly. The mother genotype is XbX, one X is unaffected, as daughters are not affected by the disease. First son has XY genotype, Second son has XbY genotype, First daughter has XbX genotype, third son has XbY genotype, second daughter has XbX genotype, fourth son has XbY genotype and fifth son has XY genotype. b. Yes. For example, AABbccDd genotype might contribute +1.3 units to genetic liability while aabbCcdd might contribute -2.3 units to genetic liability. Due to unmeasure genetic liability, any arbitrary unit can be set. It can be 0 which denote the mean liability and positive numbers denote risk and negative denote protection i.e., lower probablity of developing the disease. Polydactyly mutation was mapped to chromosome 2p. A single point mutation is maintained to carry the mutation. A study showed that five-generation Indian family of 37 individuals have 15 affected persons of polydactyly and genome wide mapping and haplotype analysis observed in this study indicated highly informative polymorphic markers on part of pedigree showed linkage between the phenotype and markers on the chromosome. c. The first daughter is likely to be carrier of this mutation as analyzed from the restriction fragments of the gene. Recombination between marker and disease can never be completely ruled out, even for very tightly linked markers, but the error rate can be greatly reduced by using two marker loci situated on opposite sides of the disease locus. This type of flanking or bridging markers help to produce a marker-marker recombination and at least false prediction can be avoided. Solution a. From the DNA fragmentation analysis, it is clear that the polydactyly came from the parents X chromosome as the affected sons have same band pattern as father has one band that found in sons and mother has two bands that appeared in affected sons. The father genotype is XbY, where Xb is affected with polydactyly. The mother genotype is XbX, one X is unaffected, as daughters are not affected by the disease. First son has XY genotype, Second son has XbY genotype, First daughter has XbX genotype, third son has XbY genotype, second daughter has XbX genotype, fourth son has XbY genotype and fifth son has XY genotype. b. Yes. For example, AABbccDd genotype might contribute +1.3 units to genetic liability while aabbCcdd might contribute -2.3 units to genetic liability. Due to unmeasure genetic liability, any arbitrary unit can be set. It can be 0 which denote the mean liability and positive numbers denote risk and negative denote protection i.e., lower probablity of developing the disease. Polydactyly mutation was mapped to chromosome 2p. A single point mutation is.

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joint distribution function of X and Y will be given by int->( infite to infinte ) int ->( infite to infinte)g(x,y)dydx int->ingration means but here the values are between 0 to 1 int->( 0 to 1)int ->( 0 to 1)(Y-X)dXdY int->( 0 to1) (0 to 1) (YX-X^2 / 2)= int ->( 0 to 1) (Y-1/2)dY=y^2/2-y=1/2-1=-1/2 Solution joint distribution function of X and Y will be given by int->( infite to infinte ) int ->( infite to infinte)g(x,y)dydx int->ingration means but here the values are between 0 to 1 int->( 0 to 1)int ->( 0 to 1)(Y-X)dXdY int->( 0 to1) (0 to 1) (YX-X^2 / 2)= int ->( 0 to 1) (Y-1/2)dY=y^2/2-y=1/2-1=-1/2.

joint distribution function of X and Y will be given by int-( i.pdf

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Answer part a)According to the gene map of the wild type and the m.pdf

1. Answer part a) According to the gene map of the wild type and the mutant gene, it can be seen that the mutant lacks a site for restriction digestion. This shows that in wild type, the restriction enzyme can digest the DNA at any three sites and thus atleast two bands will appear after amplification. On the contrary, the deletion of middle sequence of restriction site will lead to formation of a single band in mutant gene. The smaller band will represnt the mutant and the larger band will represent the normal/wild type allele. Thus, according to this information, it can be deciphered that the mother is a heterozygous in nature and carries a mutant allele for Tay-Sachs disease while the father does not. Thus, the lower band will be a representative of mutant or "t" and the upper band will be a representative of normal "T" allele. Answer part b) Since the mother is a carrier, there are 50% chances that the mother will transmit the mutant allele to her offspring. Thus, the couple can bear a normal heterozygous carrier female offspring or an affected male offspring. According to the information in the image, the genotype bands of father and offspring match which clearly suggest that the offspring is an unaffected heterozygous carrier female. Answer part c) There are always 50% changes for a male to be affected by the disease. Rest 50% males will be normal both geotypically and phenotypically. Also, all the females will be phenotypically normal but 50% of them would be heterozygous carriers. Genotypes: XtX : XtY : XX : XY :: 1:1:1:1 Phenotypes: Affected : normal : carriers :: 1: 2: 1XYXtXtXXtYXXXXY Solution Answer part a) According to the gene map of the wild type and the mutant gene, it can be seen that the mutant lacks a site for restriction digestion. This shows that in wild type, the restriction enzyme can digest the DNA at any three sites and thus atleast two bands will appear after amplification. On the contrary, the deletion of middle sequence of restriction site will lead to formation of a single band in mutant gene. The smaller band will represnt the mutant and the larger band will represent the normal/wild type allele. Thus, according to this information, it can be deciphered that the mother is a heterozygous in nature and carries a mutant allele for Tay-Sachs disease while the father does not. Thus, the lower band will be a representative of mutant or "t" and the upper band will be a representative of normal "T" allele.

2. Answer part b) Since the mother is a carrier, there are 50% chances that the mother will transmit the mutant allele to her offspring. Thus, the couple can bear a normal heterozygous carrier female offspring or an affected male offspring. According to the information in the image, the genotype bands of father and offspring match which clearly suggest that the offspring is an unaffected heterozygous carrier female. Answer part c) There are always 50% changes for a male to be affected by the disease. Rest 50% males will be normal both geotypically and phenotypically. Also, all the females will be phenotypically normal but 50% of them would be heterozygous carriers. Genotypes: XtX : XtY : XX : XY :: 1:1:1:1 Phenotypes: Affected : normal : carriers :: 1: 2: 1XYXtXtXXtYXXXXY

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