11.32 For the pedigrees A and B, indicate whether the trait involved in each case could be recessive or domi- nant, and explain your answers. Generation I II Pedigree A 1 2 1 2 3 4 III 1 2 3 4 Pedigree B Generation I 1 2 II III 1 2 3 4567 1 2 3
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- FAlpQLSfiOhfAvlhxzCSiUll_6rt-nU5b0WI73UmWOxkOw8OCwk01ng/formResponse B 1 2 Bb x Bb b 4 The fur in both parents in this cross is * 1 B B Bb x Bb b 3 4 brown black O homozygous dominant homozygous recessive 3. 近485 & :56M ******* 24 DIHYBRID CROSSES DRV 0 Stv T alı A @ zladenA 9160p2-id2 bns obidalbaneoviene da II\ MOD YR 21 $59A ... Create a dihybrid cross and determine the expected phenotypic percentages of the offspring of two corn plants both of which are heterozygous for colour and texture (RrTt X RrTt). Don't forget to include clear let statements, and follow the all six steps taught on solving genetics problems. insig moni nellog: bna. zoom
- 3.15 The pedigree shown here is for a rare autosomal recessive trait with complete penetrance. You may assume that no one in the pedigree has the recessive allele unless that person inherits it from either I-1 or II-4 or both.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 9XI II III II-1 11-2 Aa 1-1 11-3 Aa 1-2 II-4 III-1 Example: Given the pedigree above, what is the probability that the granddaughter (III-1) will be heterozygous? In order to determine the probability that III-1 is heterozygous, we need to determine the probability of the possible genotypes for her mother (II-2). Because the grandparents are both heterozygous, we would expect the following genotypic ratios in their offspring: ¼ AA, ½ Aa, and 14 aa. However, we have to use all possible information, and the circle representing III-1 is not shaded in, so she cannot be aa. Therefore, we eliminate the probability that II-2 is aa, and the final probability for II-2 is 1/3 AA and 2/3 Aa. We assume II-1 is AA because he is marrying into the family and we assume everyone marrying into a family is homozygous wild-type, unless proven otherwise. If II-2 is AA, and we assume II-1 is AA, then there is no chance that their daughter (III-1) is Aa If II-2 is Aa, and we assume II-1 is AA, then there is a…
- 1. The pedigree below shows the incidence of rare, autosomal dominant disorder called Ehlers-Danlos disease. The pedigree covers three generations of a particular family and also shows individual genotypes at a potential marker locus (M). a) Indicate the phase of all gen II and III individuals. DdM1M3 ddM2M6 II DDM3M6 ddM4M5 III DdMзM4 DdMЗМ5 DDM3M4 ddM3M5 DDM3M4 ddM5M6 DDM3M4 ddM4M6 ddM5M6 ddM5M6 b) Which, if any, of the gen III individuals are recombinants? c) Calculate the LOD score as a test of physical linkage between the marker (M) and the disease locus. d) What do you conclude about linkage between D and M?Consider the following pedigree. 하 3 10 (5 3 2 (a) What pattern of transmission is most consistent with this pedigree? (1) autosomal recessive, (2) autosomal dominant, (3) X-linked recessive, (4) X-linked dominant. (b) If individual V-2 marries a normal individual, and if the condition has a pene-trance of 85 percent, what is the probability that their second child will express the trait? (c) On the third line, what does the diamond with a 10 in the middle mean?Classes SBI3C1-2 rr x rr Meet - rz pQLSeUir31BTTSeUl8EYpVNYpajrmzBg_g0n6oMivineMfM4k0w/viewform rr x Rr Classwork O Rrx Rr ORR X Rr Genet X SBI3C1-2 Genetics Two parents were known to be right-handed. Assuming that right-handed (R) is dominant to left-handed (r), what would be the genotypes of the parents if their son is left-handed? Google M Post Atte Sp * 1 poir
- 79, 6 77 10 9I The following 5 population genetics questions refer to this introduction: "Tay-Sachs disease is inherited as an autosomal recessive, caused by the absence of hexosaminidase-A (Hex-A). Without Hex-A a lipid accumulates in cells, particularly in the brain, resulting in homozygous recessive individuals regressing in mental and physical function until death in early childhood. Heterozygotes have abnormal Hex-A activity but manifest no disease symptoms. In a large eastern European population, the frequency of Tay-Sachs disease is 3 percent. Using this information answer the following." a) If the population is assumed to be in Hardy-Weinberg equilibrium with respect to Tay-Sachs, what is the frequency of the allele that causes Tay-Sachs (answer in two decimal points [DP])? b) What would be the frequency of heterozygotes (answer in 2 DP)? Assessment Navigator c) Assuming no assortative mating, what is the probability of two heterozygotes mating (answer in 2DP)? 1 2 3 d) In…In 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.…
