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Leviton 49F44-GY 49 Series Single Pole Rhino-Hide Female Connector - Gray Leviton 49F44-GY 49 Series Single Pole Rhino-Hide Female Connector - Gray
List Price: $495.26
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Rhino-Hide 49 Series Single Pole Female Connector (Contact, Insulator, and Dust Cap), Industrial Grade, Crimped, 444 MCM Cable, 1000 Volt, 796 Amp Max - GRAY

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Linkage Single Female
Linkage Single Female

Explain these figures in terms of linkage.?

In a sort of flies, the allele b gives the body and black body + b da phenotype Brown wild type. The w allele of a gene that gives wings + w Corrugated wings give non-corrugated, wild-type phenotype. Allele g gene provides third-green eyes and red g + gives the phenotype of wild type. A woman heterozygous for these three genes is the offspring of the test Cross and 1000 are classified as follows on May 6 double-crossing black wavy green double-cross 69 black, wavy green Crossover 67 Crossover single individual (distance between B and G) 382 Wavy no crossover (original) 379 black, green Crossover No. 48 individual wild-type crossover, wavy, dark green single 44 Crossover (distance between gyw) (a) Explain these figures in terms of linkage. (B) Calculate the distance map and determine if there is interference. (C) draw alleles in their correct positions in the chromosomes of the female heterozygous triple used in the melee.

A woman + b + g- body is testcrossed + c which means that the cross is: b + c + cp + gx bbwwgg Thus, children must be an equal number of eight possible genotypes (assuming that all genes are on different chromosomes and independently order): + cp + b + g b + ca + b + g b + w WWG wwgg WGG BBW BBW Ca + + + + g + G WGG bbwwg bbwwgg However, since the numbers are skewed, indicating the link. The descriptions are recessive phenotype phenotype and indicate that genes are linked in a single chromosome. Therefore, the only thing to determine is the order and map distances. So taking the above possibilities and investment the number of flies with the genotypes with them: a + b + g + pc GT 48 + b + g + 382 pp bwwg b + b + 67 6 + BBW bwwgg WGG + ca + 5379 g + 69 g bbwwg BBW WGG bbwwgg 44 The largest number were the parental genotypes (b + g + and + WGG bwwg BBW) than those resulting from the absence of-way. This means that you 've thrown a little curve in the problem which the heterozygote has a chromosome with two alleles of wild-type and chromosome homologous allele third wild type, rather than the three wild-type alleles in the same chromosome. In the genotypes of the parents, you can say that the wild type gene is the other chromosome (w +). Thus, the chromosomes in the heterozygotes would be: b + + gw / + g wb Note that we have not yet discovered the order of genes. The double crossovers (b + + + 6 and bwwgg ca g BBW 5) in general, is to give the maximum information on the order of genes and why! The control options are: b + g + G + + + + Wow ob b gw order to have a double crossover event to give a + b bwwgg and give BBW Ca + + G, then the order must be b + g + w + w bg You can draw three possibilities on paper and make double crossovers to view. The person most common cross (A + B + w bggw 67 + 69 + b + bbg GWW bg gw + W 48, and bbggww 44) and they will help you determine the distance. The genes on the chromosomes in the heterozygous are the following: b + g + w / w + bg Give them a different way shows better chromosomes: a + b + g + bg ww Since different products are transients + b + W67 bggw bbg + 69 + b + + GWW bg gw W 48 bbggww 44, the first is a cross between B and G, and the second is the opposite. The third is the intersection gywy fourth is the opposite. As it happened more crossover between genes b and g, then there is more distance between these genes because it no more chance of crossover. The distances are: b - G (69 +67) / 1000 = 13.6% g - w (48 + 44) / 1000 = 9.2% Thus, the genetic distances are: B to G is 13.6 map units and g is 9.2 W map units.

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Written by editor

December 29th, 2002 at 7:38 am

Posted in Vintage Parts

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