The pedigree chart below indicates that individual I12 has cystic fibrosis. What is the chance that I13 is a carrier of the cystic fibrosis allele? Please explain how you determined this.
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- 1. The pedigree chart in Figure 5.29 shows the inheritance of haemopiu family. Study the pattern of inheritance in the pedigree chart, and then answer the questions that follow. о 5 6. 3 8 9 10 11 Key Unaffected male Haemophiliac male О Unaffected female Fig. 5.29 Pedigree chart of a family affected by haemophilia a) What is the genotype and phenotype of individuals 2 and 4? b) (i) How many of the unaffected family members are definitely carriers of the recessive allele? (ii) How are you able to tell which of the family members are carriers? (4) (1) (3) c) (i) If Individual 11 marries a carrier female, what percentage of their sons is likely to be haemophiliacs? (1) (ii) Use a genetic diagram to show how you worked out your answer in i, (6) 2. Why is haemophilia never passed from father to son, even though it is most common in males? (4) 3. Can a mother pass on a sex-linked gene to her daughter? (1) 4. Sipho has red-green colour blindness. One of his grandfathers was also. colour…Consider this pedigree showing an autosomal dominant rare disorder. What is the degree of penetrance? Show your work. na оп 16 19 fa 16 R 9XNeurofibromatosis-1 (NF1) is an autosomal dominant disorder where tumours form in the base layer of the skin or in nerve tissues. What is the probability that individuals II-1 and II-2 will have a genetic son with NF1? Find the image attached.
- Shown above is a family pedigree tree in which family members afflictedwith the disease Haemophilia are shown with filled-in squares (male) or circles (females). A couple is trying to determine the likelihood of passingon the disease to their future children (represented by the ? symbolabove) because the hemophilia runs in the woman’s family. Turner syndrome is a disease in which an individual is bornwith only a single X chromosome. Suppose the woman in thecouple is a carrier for hemophilia and has a child with Turnersyndrome. Would this child have the disease?The mother of a family with 10 children has blood typeRh+. She also has a very rare condition (elliptocytosis,phenotype E) that causes red blood cells to be oval rather than round in shape but that produces no adverseclinical effects. The father is Rh− (lacks the Rh+ antigen)and has normal red blood cells (phenotype e). The children are 1 Rh+ e, 4 Rh+ E, and 5 Rh− e. Information isavailable on the mother’s parents, who are Rh+ E andRh− e. One of the 10 children (who is Rh+ E) marriessomeone who is Rh+ e, and they have an Rh+ E child.a. Draw the pedigree of this whole family.b. Is the pedigree in agreement with the hypothesisthat the Rh+ allele is dominant and Rh− is recessive?c. What is the mechanism of transmission ofelliptocytosis?d. Could the genes governing the E and Rh phenotypesbe on the same chromosome? If so, estimate the mapdistance between them, and comment on your resultHere are the progeny of this cross: (Note that the categories are not in any particular order.)Fly type # of prog. Phenotype symbols Categorywt eyes black body wt wings 97 grn+ blk crv+Green eyes black body curved wings 709 ParentalGreen eyes wt body wt wings 9Green eyes black body wt wings 162wt eyes wt body wt wings 727wt eyes black body curved wings 12wt eyes wt body curved wings 179Green eyes wt body curved wings 105Total = 2000 9.) Write the phenotype symbols in the right-hand column. The first one has been done for you.10.) Next to that, label all fly categories as parental (NCOs), SCOs, and DCOs. One has been donefor you.11.) After each SCO/DCO label, write which gene got “unlinked” in these offspring.12.) Put these three genes into a genetic map in the proper order.13.) Calculate the genetic distance between the genes and label the map with these distances.14.) Calculate the cross-over interference15.) Return to questions #1-6 above. For question 6, you gave your opinion, but…
- From the powerpoint presentation, choose five (5) terms or concepts pertaining to chromosomal abnormality in the number. Describe each in a sentence. https://www.slideshare.net/farhanali911/chromosomal-abnormalities-33461290In this case a family history revealed a genetic basis for the disorder. The pedigree is shown in Fig. 1 Below. Key Ø Female: affected Female: unaffected || IV V 5600 orize 077808 15 10 9 10 CHO વ Male: affected Male: unaffected Deceased Disease status not given Dizygotic twins Monozygotic twins Fig. 1 Disease pedigree. Five generations I, II, III, IV, V are shown. Females are represented by circles, males by squares, dizygotic (non-identical) twins by diagonal lines originating from the same point, Monozygotic (identical) twins by diagonal lines originating from the same point and joined symbols and deceased by a diagonal line through the symbol. Filled symbols indicate that the individual displays the disease phenotype. Unfilled symbols indicate that the individual does not display the disease phenotype. Carriers of the disease are not indicated. Information on disease status is not known for generation I and is omitted for the individuals represented by a symbol with an asterisk.…Please consider the following pedigree. Assume that people who marry in to the family do not carry the allele unless otherwise indicated. Assume complete penetrance. I II 5 6 III 6 IV 1 2 a. Is it possible for the inheritance pattern for the trait illustrated in this pedigree to be as a result of each of the following? Answer yes or no. (i) an autosomal recessive allele (AR) (ii) an autosomal dominant allele (AD) (iii) a X-linked recessive allele (XR) (iv) a X-linked dominant allele (XD) b. Provide a genotype for individual III-6 for the most likely mode of inheritance as determined in (a).
- Consider the selfed offspring of a AaBbCcDd individual: What is the probability that offspring will have the following genotypes: AABBCCDD AaBbCCDd ● • A_B_C_D_ BbCCDd ● ● ● ● ● 1/4 x 1/4 x 1/4 x 1/4= 1/256 1/2 x 1/2 x 1/4 x 1/2= 1/32 3/4 x 3/4 x 3/4 x 3/4= 81/256 1 x 1/2 x 1/4 x 1/2= 1/16 Same genotype as the parent? A.1 B. 1/1/12 C. D.1/8 E.1/16Assume that there is complete dominance and complete penetrance at each locus and that epistasis does not occur. (These are the same conditions we used in class). Write your answers in numeric form (not in words). Referring to the cross: Dd EE GG hh xMxm x Dd ee Gg HH xMY How many different gametes are produced by the individual of genotype DdEEGGhhXMXm? What is the probability that the first offspring from this cross will be genotype DDEeGgHhXMxm? our answer rounded properly to 5 decimal digits. What is the probability that the first offspring from this cross will be a son that shows the dominant phenotype for all loci? (Type in your answer rounded properly to 5 decimal digits.)Kelly and Sam are both unaffected carriers for two autosomal recessive disorders, PKU (chromosome 12) and cystic fibrosis (chromosome 7). They are expecting a daughter. What is the probability that she will be unaffected by PKU, but effected by cystic fibrosis? O 1/16 O 3/16 O 1/2 О 3/4 O 9/16