Answer the following questions. 1. Construct a map for the genes d,e,f. Assume that: d and e = 3%; e and f = 5%. Give 2 %3D arrangements of the genes/maps. 2. If d and f = 2%, what is the correct arrangement of the genes d,e,f? 3. Consider the fourth gene "g". if g and e = 1.5%, give two possible arrangements. 4. If d and g = 1.5 % give the correct order of the four genes
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- Southern Blotting & Detection of sickle cell disease: Please interrupt the following results and draw a Punnett square, and answer the questions. •Discuss figures (gel figure and Punnett square) one by one. •You must provide discussion for each lane of the gel, how you make out normal, carrier, disease (allelic patterns- why so?), control IMPORTANT POINTS TO REMEMBER: •Homozygous (one band)- the size of the band (lower or higher in the gel) determines if normal (both alleles cut by MSt II, hence band lower) or diseased (both alleles uncut and hence at the higher position in gel) •Heterozygous (two bands in the same lane) denotes a carrier (lower band from normal allele which is cut by RE & upper sickle cell allele band is uncut)Answer the following questions given the pedigree below. Please assume that no other mutations are occurring, complete penetrance, and that the individual marked with an asterisk (*) doesn’t carry the allele causing the affected phenotype. Q2) What are the genotypes of the following individuals listed in the table below. Use the uppercase “A” to represent the dominant allele and lowercase “a” for the recessive allele. Individual All possible genotypes I-2 II-1 IV-2 V-2 II-2 II-31. Construct a map for the genes d,e,f. Assume that: d and e = 3%; e and f = 5%. Give 2 arrangements of the genes/maps. 2. If d and f = 2%, what is the correct arrangement of the genes d,e,f? 3. Consider the fourth gene “g”. if g and e = 1.5%, give two possible arrangements. 4. If d and g = 1.5 % give the correct order of the four genes
- 8. Huntington’s disease is a degenerative disease of the nervous system that strikes in middle age. The allele that causes the disease (H) is dominant to the allele that results in the normal condition (h). Answer the following questions about the inheritance of this disease. A. What is the genotype of a man who is normal but whose father had Huntington’s disease? B. What is the genotype of a woman who has Huntington’s disease if both of her parents had Huntington’s disease? C. If a man who is heterozygous for Huntington’s disease marries a woman who is normal, what would you expect for the genotypes and phenotypes of their children? D. If a normal man marries a woman who is homozygous for Huntington’s disease, what do you expect for the genotypes and phenotypes of their children?Answer the following questions given the pedigree below. Please assume that no other mutations are occurring, complete penetrance, and that the individual marked with an asterisk (*) doesn’t carry the allele causing the affected phenotype. Q4) Assuming II-2 and II-3 want to have another child. what are all the possible genotypes, and what is the percentage that their child will be affected?1. The first genetic test used a short DNA sequence that was closely linked to the Huntington’s locus. The linkage distance between the DNA sequence used in the test and the actual Huntington’ disease locus was 5%. a. In a large sample of the Venezuelan family (over 2,000 individuals), what percentage of people would inherit the DNA sequence but not the Huntington’s allele (i.e. be a false positive test for Huntington’s)? b. What percentage of that large Venezuelan family would not inherit the DNA sequence but would inherit the Huntington’s allele (i.e., be a false negative test for Huntington’s)?
- Explore chromosomal defects among humans in Ensembl at https://www.ensembl.org/index.html?redirect=no(Links to an external site.) Take a screenshot of one of your findings and describe the type of aberration and its associated phenotype. Identify the locus of gene variant(s) responsible for the disorder. Screenshot Name of chromosomal defect Type of aberration and locus of gene variant(s) Phenotype References:Briefly discuss why mutant allele 1 fails to produce functional protein. (include three points in your answer)People are characterized by abnormal red patches, irritated skin, eczema, decreased number of platelets and bloody diarrhea.1. Name the trait that causes these symptoms. 2. If this is a heritable trait, what is the mode of inheritance?3. What is the type of mutation cause this syndrome?
- Construct a map for the genes d,e,f. Assume that: d and e = 3%; e and f = 5%. Give 2arrangements of the genes/maps.Complete the sentences You cross two mice with wildtype (long) tails and collect a male offspring with a short tail, who you name Squeak. Tail length is controlled by the gene Lf, which is a paternally imprinted gene in mice. Given that this is an imprinted gene, explain the inheritance and expression of the trait. 1. Squeak’s short tail allele was inherited from the _____ (female, male, or either) parent. That parent did not express the trait because the allele was ____ (unmethylated, methylated) in the parent’s somatic cells and was inherited from that parent’s own ___ (mother, father, either mother or father)2. If Squeak’s offspring inherit the short tail allele from Squeak, will they have short tails? Yes or No?3. If Squeak was a female instead of a male, which answers in this question would change? ____ (some of the answers, all of the answers, or only the one about the offspring)Alpha thalassemia is a hereditary blood condition that results in varying levels of anemia. It is tied to the HB alpha 1 gene and the HB alpha 2 gene on human chromosome 16. The diagram shows the proteins for the hemoglobin genes and the pedigree shows genotypes, designated by the letter X, on the chromosomes for a family affected by the condition. Which represents the predicted level of anemia in a child born to the mother and father in the image with a mutation that results in a genotype of xxxx? Why? A - mild anemia because the loss of 4 genes would equal the loss of the 4 proteins needed for normal alpha hemoglobin B - severe anemia because the loss of 4 genes would equal the loss of the 4 proteins needed for normal alpha hemoglobin C - mild anemia because the addition of 4 genes would produce too many of the proteins needed for normal alpha hemoglobin D - severe anemia because the addition of 4 genes would produce too many of the proteins needed for normal alpha hemoglobin