5. The following image is of ASOanalysis performed at 90°C on afamily with Ty Sachs (M and F are theparents, C1, C2 and C3 are thechildren), and 2 unrelated controls.These parents had a normal child andthen had a child with Ty Sachs andthey want to know if their third childwill have it. The top ASO dotrepresents the Ty Sachs allele andthe bottom represents the normalallele.b. What is the status of the 3 children?c. What are the controls?TZNa. What is the status of the parents? (Have disease, are carriers, are normal?)d. The following panel is of a similar analysisdone, excpet the technician found that themachine had been set to the wrongtemperature for analysis. Is the temperaturethe machine set to higher or lower? Explainwhy.6. A murder victim is found in an advanced state ofdecomposition and cannot be identified. Policesuspect that the victim is one of 5 individualsreported missing by their parents. DNA typing(RFLP) is carried out against those 5 sets ofparents.According to loci 1 only, which couple could be theparents of the dead individual found?Who are the parents of the individual found(combine both data)?According to loci 2 only, which couple could be theparents of the dead individual found?MΤΖFNC1C1A B C D Ed d d d d d 59 59 xControl 1-Control 1Control 2Control 2A B C D E6 9 6 2 d 98989 X

Name: 5. The following image is of ASOanalysis performed at 90°C on afamily
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Description: 5. The following image is of ASOanalysis performed at 90°C on afamily with Ty Sachs (M and F are theparents, C1, C2 and C3 are thechildren), and 2 unrelated controls.These parents had a normal child andthen had a child with Ty Sachs andthey want to

Tay-Sachs disease is a rare genetic metabolic disease that primarily affects young children. It is…

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Biology

5. The following image is of ASO analysis performed at 90°C on a family with Ty Sachs (M and F are the parents, C1, C2 and C3 are the children), and 2 unrelated controls. These parents had a normal child and then had a child with Ty Sachs and they want to know if their third child will have it. The top ASO dot represents the Ty Sachs allele and the bottom represents the normal allele. TZ N M F C1 Control 1 Control 2

5. The following image is of ASO analysis performed at 90°C on a family with Ty Sachs (M and F are the parents, C1, C2 and C3 are the children), and 2 unrelated controls. These parents had a normal child and then had a child with Ty Sachs and they want to know if their third child will have it. The top ASO dot represents the Ty Sachs allele and the bottom represents the normal allele. TZ N M F C1 Control 1 Control 2

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter12: Mendel, Genes, And Inheritance

Section: Chapter Questions

Problem 7TYK: In cats, the genotype AA produces tabby fur color; Aa is also a tabby, and aa is black. Another gene...

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1. A mother with blood type A has a child with blood type O, and she claims that Mister X is the father. Mister X denies that he could possibly be the father, because he has blood type B. If you were the judge presiding over this case, which of the following arguments would you find most convincing? a. Mr. X could be the father if his blood genotype is IBIB, and the mother’s blood genotype is IAIA b. Mr. X cannot be the father because a child with blood type O cannot have a parent with blood type B c. Mr. X could be the father if his blood genotype is IBi, and the mother’s blood genotype is IAi d. Mr. X cannot be the father because a parent with blood type B cannot have a child with blood type O e. Mr. X cannot be the father because the human ABO blood types are sex-linked traits 2. A monohybrid cross between two pea plants producing yellow peas, resulting in an F1 phenotypic ratio of 3:1 (three yellow pea plants to one green pea plant) indicates what about the parental pea plants?…
3. A couple just had a baby, who has type-O blood. Upon learning the baby’s blood type, the father becomesinfuriated and accuses the mother of cheating on him. The mother has type-A blood and the father has type-Bblood. Knowing that the allele for type-O is recessive, answer the following questions.a. What is the relationship between the IA, IB, and i alleles?b. Is the father right? Or is he falsely accusing the mother? Explain. Include the genotypes of the parentsand the genotypes of their potential children.
1. a) If you were at risk of a fatal disease such as Huntington’s disease and only the 95% accurate preliminary genetic test was available, would you take the test? After all, 95% of the time the test would tell you whether you would eventually get Huntington’s disease or whether you did not have to worry about getting this disease. b) The linkage distance between the DNA sequence used in the test and the actual Huntington’ disease locus was 5%. In a large sample of the Venezuelan family, 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)? 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)?
2.In dogs, there is a hereditary deafness caused by a recessive allele, d. A kennel owner has a male dog that she wants to use for breeding purposes if possible. The dog can hear, but the owner is unsure of the genotype. She does a testcross (crosses it to a homozygous recessive dog), and two of the five offspring are deaf. This means that the male dog A.has the genotype DD B.has the genotype Dd C.is still of unknown genotype since there were offspring of both deaf and hearing phenotypes. D.has the genotype dd
1. A mutant allele in persons with familial hypercholesterolemia (FH) causes death due to a lack of liver receptors for LDL. Susceptible persons have half the normal number of receptors, while other individuals have the normal number of receptors. In a phenotypically normal couple, the man had a female 1st cousin (on his father’s side) who died from FH; the woman had a maternal uncle with FH. Using standard symbols, draw a pedigree of these couple’s families. Write the genotypes of each family member. Indicate your legend for symbols used.
4. Mr. Salazar has Type A blood while Mrs. Salazar has Type B. They have four children. Bobbie has Type A, Gabbie has Type B, Alex has Type AB, and Teddie has Type O. What are the genotypes of all six people in this family? NOTE: The ABO blood type gene has three alleles. IA and IB are codominant; i (for Type O) is recessive to both.
For each genotype listed, what allele combinations will occur in gametes? a. AABBCC b. AaBBc c. AaBBcc d. AaBbCc
In cats, the genotype AA produces tabby fur color; Aa is also a tabby, and aa is black. Another gene at a different locus is epistatic to the gene for fur color. When present in its dominant W form (WW or Ww), this gene blocks the formation of fur color and all the offspring are white; ww individuals develop normal fur color. What fur colors, and in what proportions, would you expect from the cross AaWw Aa Ww?
A single allele gives rise to the Hbs form of hemoglobin. Individuals who are homozygous for the allele (HbS/HbS) develop sickle-cell anemia (Section 9.6). Heterozygous individuals (HbA/HbS) have few symptoms. A couple who are both heterozygous for the HbS allele plan to have children. For each of the pregnancies, state the probability that they will have a child who is: a. homozygous for the HbS allele b. homozygous for the normal allele (HbA) c. heterozygous: HbA/HbS
A heterozygous individual has a _______ for a trait being studied. a. pair of identical alleles b. pair of nonidentical alleles c. haploid condition, in genetic terms
A testcross is a way to determine ________. a. phenotype b. genotype c. dominance
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)?