Consider the following cross examining four gene in two parental lines: Parent 1: A/a; B/b; D/d; e/e Parent 2: a/a; b/b; d/d; Ele Assuming independent assortment for the four genes, what fraction of progeny will have the same genotype as that of parent 2? 1/2 O 1/4 O 1/8 O 1/16
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- Consider the following cross examining four gene in two parental line: Parent 1: A/a; B/B; D/d; E/e Parent 2: A/a; B/b; d/d; e/e Assuming independent assortment for the four genes, what fraction of progeny will be phenotypically identical to parent 2? O 3/32 O 3/16 none of the above 3/8 O 3/4Consider the following cross examining four gene in two parental line: Parent 1: A/a; B/B; D/d; E/e Parent 2: A/a; B/b; d/d; e/e Assuming independent assortment for the four genes, what fraction of progeny will be phenotypically identical to either parent 1 or parent 2? (Hint: first figure out the fraction of progeny that resembles parent 1 and parent 2 separately, then get the overall fraction.) 9/16 1/16 3/4 3/8 3/16Calculate the probability of either all-dominant or all-recessive genotypes for the alleles A, B, E, and F in the following cross: A/a;B/b;c/c;d/d;E/e;F/fx A/a;B/b;C/c;d/d;E/e;F/f 1/64 1/128 1/256 1/32 1/16
- An individual heterozygous for four genes, A/a • B/b •C/c • D/d, is testcrossed with a/a • b/b • c/c • d/d, and 1000progeny are classified by the gametic contribution ofthe heterozygous parent as follows:a • B • C • D 42A • b • c • d 43A • B • C • d 140a • b • c • D 145a • B • c • D 6A • b • C • d 9A • B • c • d 305a • b • C • D 310a. Which genes are linked?b. If two pure-breeding lines had been crossed toproduce the heterozygous individual, what would theirgenotypes have been?c. Draw a linkage map of the linked genes, showing theorder and the distances in map units.d. Calculate an interference value, if appropriateIn 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.…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.1/16
- F1 hybrid is obtained from the following cross: Parent 1: AB/AB; d/d; E/E Parent 2: ab/ab; D/D; e/e Assume that all genes are sorted independently except for genes A and B, which are tightly linked and show zero recombination. Following a testcross of F1, what proportion of the F2 progeny will be phenotypically like parent 1? O 1/32 1/8 O 1/16 O 1/4 O9/64For each genotype, indicate whether it is heterozygous (HE) or homozygous (HO) AA Bb Cc Dd Ee ff GG HH PP Pp pp Round seeds are dominant to wrinkled RR Rr rr Ii Jj kk For each of the genotypes below, determine the phenotype. Purple flowers are dominan to white flowers L1 Mm nn straight straight curly 00 Pp Brown eyes are domin BB Bb bb to blue eyes Bobtails are recessive (long tails dominant) TT Tt tt For each phenotype, list the genotypes. (Remember to use the letter of the dominant trait) Straight hair is dominant to curly. Pointed heads are dominant to round heads. pointed pointed roundIn sesame plants the one-pod condition is dominant (P) to the three-pod condition (p), and a normal leaf (L) is dominant to the wrinkled leaf (l). Pod type and leaf type are inherited independently. Determine the genotypes for the parent producing offspring: 318 one-pod, normal leaf and 98 one-pod, wrinkled leaf PPLl x PPLl, PPLl x PpLl, or PPLl x ppLl ppll x ppLl PPLL x PPLL ppll x ppll
- к, х Ааv AdBbCc AaBbCc No Spacing AaBbCc 三、相、 1 Normal Heading 1 Font Paragraph 6. Let's assume that the non-taster, daughter in question 5 is a carrier of the albino trait. She marries a Styles taster man, normally pigmented, whose mother was a non-taster albino. Show the cross between these parents. Genotype of the woman Genotype of the man Genotypes Phenotypes What is the chance that their child will be a taster? Footer What is the chance that their child will be albino? What is the chance that their child will be a taster albino? J.S.)Below is the linkage map for genes y, sh and c that are in cis arrangement. Interference is 0.4 and the total progeny is 2000. Determine the parentals, SCOI, SCO II, and DCO products and the estimated number of individuals for each. y-------------sh-- 18% 28%Considering the Mendelian traits round versus wrinkledand yellow versus green, consider the crosses belowand determine the genotypes of the parental plants byanalyzing the phenotypes of their offspring.Parental Plants Offspring(a) round, yellow * round, yellow 3/4 round, yellow1/4 wrinkled, yellow(b) wrinkled, yellow * round, yellow 6/16 wrinkled, yellow2/16 wrinkled, green6/16 round, yellow2/16 round, green(c) round, yellow * round, yellow 9/16 round, yellow3/16 round, green3/16 wrinkled, yellow1/16 wrinkled, green(d) round, yellow * wrinkled, green 1/4 round, yellow1/4 round, green1/4 wrinkled, yellow1/4 wrinkled, green