1. Perform a monohybrid mendelian cross using seed colour as one of the characteristic.Show the results at the end of the F2 generation.
2. Perform a dihybrid mendelian cross using height and colour of the flower as the two characters. Show the results at the end of F2 generation.Use Punnet square.
3. Show the probability of getting a male child during fertilisation in humans.
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Answers
Answer:
The P (Parental) cross is between true-breeding lines of wrinkled yellow peas (rrYY) and round green peas (RRyy). The F1 offspring are therefore all RrYy, and are all round and yellow. In forming the F2 plants, the alleles at the two loci segregate independently. That is, the chance of getting an R allele and a Y allele is 1/2 x 1/2, of getting an R and a y 1/2 x 1/2, and so on. Thus, all four possible diallelic combinations occur with an equal probability of 1/4. The same is true for both parents. Given four possible gamete types in each parent, there are 4 x 4 = 16 possible F2 combinations, and the probability of any particular dihybrid type is 1/4 x 1/4 = 1/16. The phenotypes and phenotypic ratios of these 16 genotype can be determined by inspection of the diagram above, called a Punnet Square after the geneticist who first used it.
Alternatively, recall that the phenotypic ratio expected for either character is 3:1, either 3 "Y" : 1 "y", or 3 "R" : 1 "R". Then, the expected phenotypic ratios of the two traits together can be calculated algebraically as a binomial distribution:
(3Y + 1y) x (3R + 1r) = 9YR + 3Yr + 3Ry + 1 ry
That is, we expect a characteristic 9:3:3:1 phenotypic ratio of round-yellow : wrinkled-yellow : round-green : wrinkled-green pea seeds.
To predict the genotypic ratios, recall that for each gene the ratio is 1 : 2 : 1 :: AA : Aa : aa . Then, algebraically
(1YY + 2Yy + 1yy) x (1RR + 2Rr + 1rr) = 1 YYRR + 2 YYRr + 1 YYrr + 2YyRR + 4YyRr + 2 yyRR + 1yyRR + 2yyRr + 1yyrr
Answer:
1) All the haploid sperm and eggs produced by meiosis received one chromosome 7. All the zygotes received one R allele (from the round parent) and one r allele (from the wrinkled parent). Because the round trait is dominant, the phenotype of all the seeds was round.