Question

Three gene loci, A, B, and D assort independently, and alleles A, B and D are dominant over alleles a, b, and d. What is the probability of An abd phenotype from a cross aabbdd x aaBbDd?

a) 1

b) 1/4

c) 1/8

d) 1/12

e) 1/64

Answer 1

Answer in are probo predi to since the alleles segregate from one another and they e independently assorted of one another robability calculation in the most efficient way then. mathematically the outcomes of netic crosses . Now, the question says, AB; and D are the dominant alle a, b, and d are the recessive and all the genes show independent assortment so as they show independent assortment, we con first calculate the probability by comes I sidering the gene loci for each character The given cross - I aa bb dd x a a Bb Dd. we fest first consider cross for gene A- a a Here both the allele is a . probability for pairing of a in each both chromosome iniz 1x1 aa aa aa aa 1 Resulting Probabile Now 10 bb x' Bb L L Probability of Probability of getting b' allele B = 7 gettin forming Resulting A probability of bb genotype. - 1x 12

Now coming to Digene, dd x Dd. . Probability of Probability d allele in gainele getting the getting a allele in geumele genotype . Resulling probability of dd azalx19) Now we have to calculate the total probability of the resulting : phenotype for 3 hansa genes i aa bis ad - = 1 x 1 2 x An Ans Ans - 1 is the resulting probability A for the habebitos abd as asked in the question desired) phenotype. for

The method of predicting the outcomes of genetic crosses is very popular using the prniciples of probability.It gives us a quantitative value of the result of crosses. Although Punnett squares are an effective and straightforward way to examine simple genetic crosses, they are unwieldy and time consuming when the inheritance of multiple characters is examined. A more effective way to examine inheritance is to use two basic laws of probability (The Rule of Multiplication and The Rule of Addition) to determine the probabilities of different outcomes when two individuals are crossed .So this mathod of calculation using the laws of probability is more efficient in crosses involving more than one characters as explained above in the picture.