children (Sex unspecified) from these parents, what proportion is he questions.

Question

children (Sex unspecified) from these parents, what proportion is he questions. Y HINT: it will be easiest to first draw a pedigree and then ans also need to use Punnet squares. 7. . The inhe eritance of coat colors of cattle involves a multiple allelic series with a as follows: S>sh. The S allele puts a band of white color around the middle of the animal and is referred to as a Dutch belt; the sh allele Hereford type spotting: solid color is the result of the se allele; and produces Holstein-type spotting e s allele. Homozygous Dutch belted-males are crossed to Holstein-type les. The Fl females are crossed to a Hereford-type spotted male of genotype shsc. Predict the genotypes and phenotypes of the progeny. n cats, short hair is dominant over long hair; the gene is autosomal. Another gene, BI, which is sexlinked,produces yellow coat color, its allele B2 produces black coat color and the heterozygous combination Bl/B2 produces tortoise shell coat color. If a long haired black male is mated with a tortoise shell female heterozy gous for short hair, what are the genotypes of the kittens produced in the F1? If the FI cats are allowed to interbreed freely, what are the chances of obtaining a long-haired yellow male? 8)l 9)Two recessive genes in Drosophilia (b and vg) produce black body and vestigial wings respectively. When wildtype flies are testcrossed the F1 all have the dominant genes on one chromosome and the recessive genes on the homologous chromosome. Testerossing the female Fl produced 1888 black and vestigial 1930 wildtype 412 black 370 vestigial (a) Calculate the distance between b and vg (b) Another recessive gene cn lies between the loci of b and vg producing cinnabar eye color. When wildtype flies are testcrossed the FI are all trihybrid. Testcrossing the Fl females produced 72 black, cinnabar 664 wildtype 68 vestigial 4 black, vestigial 652 black, cinnabar, vestigial 70 black 61 cinnabar, vestigial 8 cinnabar (c) Calculate the map distances experiments? Explain (d) do the b-vg distances coincide between the two 10)Consider the genetics of the ABO blood groups. A man of blood gropu B is being sued by a woman of blood group A for paternity. The woman's child is blood group O (a) is this man the father of this child? Explain. (b) If this man actually is the father of this child, specify the genotypes of both parents. (c) If it was impossible for this group man to be the father of a type O child, regardless of the mother's genotype, specify his genotype. (d) If a man was blood group AB, could he be the father of a group O child?

Explanation / Answer

Ans-7 according to question

genotype of Dutch belted male =SS

Genotype of holstein type female = ss

So F1 will be = SS,Ss,Ss, and ss

So when we cross F1(Ss) female with should(Hereford type)

Phenotype = S sh = Dutch belted type

s sh = Hereford type spotted

So there will be 50:50 Dutch belted type and Hereford type.

sh sh S S sh S sh s s sh s sh

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