Question

2. What is the probability of picking a 5 OR a king out of a deck of 52 playing cards? 3. What is the probability that offspring from a monohybrid cross for purple flower (Pp) is a heterozygote? 4. In a cross between AaBb Cc x AaBBCC, what is the probability that we would get either AABCc OR AABBCC? Activity #6 - Probability laws to solve complex genetic problems 1. Combine both the multiplication and addition laws P-purple flower, p - white flower Y- yellow seed, y-green seed R-round seed, r-wrinkled seed What is the probability that a trihybrid cross of PpYyRr would result in offspring that has purple flowers, yellow seeds, and wrinkled seeds? **Hint: OR = genotype AND = phenotype

Answer 1

2) In a set of 52 cards a total 4 suits are there nameley spades, hearts, diamonds and clubs. each suit has 13 cards ( ace, king, queen, jack , 10,9,8,7,6,5,4,3,2). This is the basic we need to know while answering probability questions regarding cards.

in this case, we have to find out the probability of picking a 5 or a king out of this 52 cards. now we know that in a set of 52 cards a total of four 5s and four kings are there.

so the probability of picking a 5 = 4/52

similarly, probability of picking a king = 4/52

therefore the probability of picking a 5 or a king =  

4/52 +4452 = 8/52

SUM RULE OF PROBABILITY-  It is for calculating the probability of either of two events that cannot occur simultaneously

PRODUCT RULE OF PROBABILITY- It is for calculating the probability of both of two events that are independent

EVENT- the set outcomes of an experiment

A In a cross between AaBb Ce x Aa BBCC; what is the probability that we would get either AABBCC00 AABbce ? Anguer) ble need either AABB CC or AABbce. In both AA is present. So first check probability of getting AA by punnett square with gametes from parents. А а A AA Aå from each parent (A.) and (1) gametes with segregate independently. So here probability to a aa aa get AA is = 1 Similarly check Square B probability to get BB and Bbasing punnet B BB z from one parent Bb= from another parent Bb. So, probability of getting BB and Bb is one since all boxes possess them 6 Bb Also, probability to get CC and Cc ? wono dal c/oclcc CC = from one parent Ce = from another parent. So, probability to get CC and Cc is one again since all boxes possess them. Therefore probability to get either AABBCC offsprings will be = 4 x X 1 = 1 or A AB b Cc. (Applying product
for problem number 3,4 and 6 please refer the images
vonen 3) What is the probability that offspring from a monohybrid couss for purple flower (Pp) is a beterozygote? Ansuer) Lets start the monehybrid cross with the parents Pp (male) x pp (female) it on Parents : P. P. x. pp . Gametes : XUD NOT A. neration o I generation : - no a P (peuple flower) Lets self pollmate the Fe Offsprings in F2 generation of Pp Gupple) x P p (people). gameles : 0 x ③ D Lets draw a funnet square for easy understanding. na Pop P PP puuple) PP (pusplay of Pp Qurple) pr (White) So, probability for people flower with heterozygde is +2 =