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In a diploid plant species, an
a. What is the order of these three linked genes?
b. Calculate the recombination fractions between eachpair of genes.
c. Why is the recombination fraction for the outside pair of genes not equal to the sum of recombination fractions between the adjacent gene pairs?
d. What is the interference value for this data set?
e. Explain the meaning of this
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Chapter 5 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
- Here 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…arrow_forwardHere is a pedigree from a family with a high prevalence of breast cancer. dx means the age at diagnosis. The letters A-E are haplotypes (alleles). dx51 [D-] dx 46 BC ☑ 7 dx58 BD (C-) 77 CD dx 44 BD AC हे dx 48 42 dx38 54 dx33 BC BC BD DA BC d.) Are there any individuals in the pedigree you might expect to have breast cancer, but do not? If so, identify the people (using the standard notation for identifying people on a pedigree) and suggest an explanation.arrow_forwardIn a diploid plant species, an F1 with the genotype Mm Rr Ss is test crossed to a pure breeding recessive plant with the genotype mm rr ss. The offspring genotypes are as follows: Genotype Number Mm Rr Ss 687 Mm Rr ss 5 Mm rr Ss 68 Mm rr ss 196 mm Rr Ss 185 mm Rr ss 72 mm rr Ss 8 mm rr ss 679 Total 1900 1. What is the gene order of these linked genes?arrow_forward
- In crosses between mutant mice, you find the following patterns in the genotype for the A locus containing alleles A1, A2, A3 and A4: What would be the best explanation for the inheritance and why?arrow_forwardThe trihybrid e+f+g+/efg is test crossed to the triple recessive efg/efg and the following genotypes were obtained in the progeny: efg = 2; efg = 28; e'fg* = 9; e*rg 32; efg 73; e*fg* = 10; ef g* =10; e'fg 2; %3D "The odd gene is The gene order is The gene distance between e-f is The gene distance between f-g is The gene distance between e-g isarrow_forwardAn individual heterozygous for d e f (D/d E/e F/f) was crossed to a homozygous recessive individual (d/d e/e f/f). The following offspring were recovered.d e + 39+ e f 372d + f 7+ + f 47+ e + 5d + + 412What is the correct order of these genes? EDF DFE DEFarrow_forward
- three recessive genes a, b, and c in the model plant Arabidopsis are found to be linked on chromosome 4. A three point test cross is done with a homozygous recessive plant with a heterozygous for all three genes. Following is the number of progenies a b C 65 A B c 56 A B C 1267 a b c 1310 A b C 550 a B c 515 a B C 470 A b c 489 Total = 4,722 Determine the middle locus by your choice of method and after that calculate the map distance between the genes in map unit (m.u.).arrow_forwardAn individual is heterozygous for a reciprocal translocation, with the following chromosomes: A • B C D E F A • B C V W X R ST • U D E F R ST • U V W X Q. Explain why the fertility of this individual is likely to be less than the fertility of an individual without a translocation.arrow_forwardConsider 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.1/16arrow_forward
- Which of the pedigree diagrams below is most likely to show a family with X-linked agammaglobulinemia? A KEY. C Horygon Hamocyga Hezexygess Hey Wild Type fursale Mek Parmak Mais Note: Completely red symbol denotes an individual exhibiting the phenotype of interest H INV V 1/4 1/2 1/2 1.0. 1/2 1/2 1/4 Typ female 1/2 B Affected Known carrier Normal Moral D O Toarrow_forwardWould appreciate help with this question please include explanation tyarrow_forwardThe result of a test cross of a yeast cell that is heterozygous for three genes are summarized in the table (below), which shows the number of offspring that inherited each combination of alleles from the heterozygous parent. Use the data collected to make a genetic map of the three genes L, Q, and D. • D-Q: o D-L: | o Q-L: | - L L I L L cM CM Alleles CM Ơ Ơ o Q o d D D d d d # 104 27 78 Report your results by entering the genetic distance between each pair of genes in the blanks below. Round to the nearest 0.1 cm. 64 28 62 106 81arrow_forward
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