a. What is the mode of transmission of this disease? b. Which SNP allele (ASO1 or ASO2) is originally linked to the disease gene? c. Draw a diagram of the event that gave rise to the genotype of individual IV-5.
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- Can someone show me why this is correct? I will like to see the possible genotypes and phenotypes together that proves the answer.I just want an explanation of what is in bold please. Lab Introduction: A dihybrid cross is a cross between individuals that involves two pairs of contrasting traits. To Predict the results of a dihybrid, cross all possible combinations of the four alleles from each parent must be considered. You will examine a dihybrid cross involving both color and texture. Purple (P), is dominate to yellow (p), and smooth texture (S) is dominant to wrinkled (s). Both parent plants are heterozygous for both traits. Review genetics and the use of Punnett squares in a biology text before doing this experiment.MATERIALS: Assume you have ear of Corn. You need a heterozygous X heterozygous 9:3:3:1, purple/yellow, starchy/sweet. PROCEDURE: From above please write out: The crop The parental (P) cross phenotype, genotype, gametes The F1 progeny Genotype and Phenotype Cross between two F1 Selfed testcross The F1 gametes The expected F2 results, genotype, phenotype, genotypic ratio, phenotypic ratio. 1.…ABO Blood TypeThe following pedigree shows the incidence of ABO blood types in a family. dentify the genotypes of the following individuals: Individual Genotype II-1 II-2 II-4 II-5 III-2 III-3
- My Question is what is the probability their first child will have hemophilia and drawn pedigrees for family members with genotypes. My explantion so far: A man has both X and Y chromosomes as sex chromosomes in his body. Here, though the brother of the man is hemophiliac, a man can’t be a carrier of hemophilia. So, it can be said that his chromosome is “XnY”.Here, the “n” stands for “normal”.Though the paternal uncle is hemophiliac, a man cannot be a carrier of hemophilia, his niece will not be a career. So it can be said that the woman is also not a carrier and has the “XnX” chromosome.So, as the mother is not a carrier, their first child does not have a chance of having hemophilia. This can be determined as it is known that there is no hidden carrier of hemophilia in the family.A RFLP is discovered that is linked to the gene for Duchenne’s muscular dystrophy (DMD). DMD is an X-linked, recessive trait. The RFLP is 2 map units from the gene for DMD. Consider the following pedigree and Southern blot using a probe that hybridizes to the RFLP. Which band/s is/are associated with DMD? What is the genotype for individuals 3 and 4? (Remember, this is an X linked disease, so use X’s and Y’s to denote). Individual 9 married a man who does NOT have muscular dystrophy, and she is pregnant. DMD is an X-linked trait. What is the probability for their child to have DMD? An amniocentesis is performed and it is determined that 9’s child in utero has only a 10 kb band that hybridizes to the same probe used above. What can you say about the child now?Construct pedigree charts using the inheritance of hemophilia in figure 92 (page 113). This is X-linked inheritance so you are required to label XX for females and XY for males. The gene responsible for the trait is represented by the superscript which should be specified in the legend.
- Below in Figure 1 is a pedigree for a family afflicted by a genetic disorder. In some populations, Cystic fibrosis has an incidence of 1 in 2500 newborns. The carrier frequency calculated from this is 1/25. Analyze the pedigree below assuming the disease is similar to cystic fibrosis in incidence and carrier frequency. However this disease may not have the same type of genetic transmission as cystic fibrosis. Assuming her father is known to NOT be a carrier, calculate the probability that IV1 is a carrier for disease. Use the Χ2 test to determine whether your proposed transmission fits this data.Help me create a pedigree of this information: Pedigree analysis: Generation 1: Normal parents (AA x AA) Generation 2: Carrier parents (AA x AS) Generation 3: Affected child (AS x AS) Generation 4: Affected grandchild (SS) This pedigree has two normal parents in the first generation. Second generation carriers carry the sickle cell trait from one parent. The disease is 25% more likely to be inherited in the third generation if both parents have the 'S' allele. If both parents have the 'S' allele, their children will have sickle cell anemia in the fourth generationTwo families are genotyped at a marker with alleles Mand m. Genotypes at a disease gene, with alleles D and d, are inferred from the larger pedigrees (not shown). The genotype of each individual is displayed below. The phase of each individual is indicated with a horizontal bar. One chromosome has the combination of alleles above the bar and the other chromosome has the combination of alleles below the bar. Use this information to complete the following tasks: A. Labeling the children as 1, 2, and 3 moving from left to right along the chart, indicate whether each child's genotype arose from a recombinant or non-recombinant gamete from their heterozygous parent. B. Write down the formula for the combined odds ratio across the two families for linkage vs. no linkage as a function of the recombination frequency (r). You do not need to simplify your expression. Family #1 MD m d m d ma md Family #2 MD MD ML D Md md MD
- n the space below, use colored pencils to create a pedigree with the following information. Follow the guidelines for a pedigree when creating this one. Ray and Elaine were married in 1970. They both had normal vision. They had 2 daughters and then a son. Both daughters, Alicia and Candace, had normal vision and never had any children of their own. The son, Mike, was colorblind. The son married Beth who also had normal vision and they had 2 children of their own, first Greg then Victoria. Victoria was colorblind, but Greg was not. Colorblindness is a sex-linked recessive trait. Do not forget what shapes are male and female. Place the names and genotypes of the people under their shape. Color your individuals the following: Red- for colorblindness White-for regular vision Blue- for individuals with regular vision but are carriers Green- unknown genotypePedigree Analysis Is a Basic Method in Human Genetics Using the pedigree provided, answer the following questions. a. Is the proband male or female? b. Is the grandfather of the proband affected? c. How many siblings does the proband have, and where is he or she in the birth order?Given the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the phenotype of this individual be?