Title | Practice Questions for using the rules of multiplication and addition in genetic crosses |
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Author | Anonymous User |
Course | Test Bank for Global Health 101 2nd edition by Richard Skolnik |
Institution | Qafqaz Universiteti |
Pages | 2 |
File Size | 72.8 KB |
File Type | |
Total Downloads | 26 |
Total Views | 192 |
Download Practice Questions for using the rules of multiplication and addition in genetic crosses PDF
Practice questions for using the rules of multiplication and addition in genetic crosses Q - When do we us the rule of multiplication? A - When we have independent events in sequence. You multiply the probabilities of the independent sequential events. Example: In heterozygous pea plants (Pp) what is the probability that the offspring (Pp xPp) will be homozygous recessive? Pp x Pp The chance of the first parent (Pp) passing on p (little p) is ½ and the second parent (Pp) is also ½. Therefore ½ x ½ = ¼ OK that is simple. In a trihybridcross between AABbCC X AaBbCc what is the probability that the offspring will be AaBbCC? Just work on each letter combination at a time. You can use a punnett square initially but you will soon learn the probabilities. AABbCC X AaBbCc resulting in AaBbCC so Aa =2/4 or ½
A a
A AA Aa
A AA Aa
Bb X Bb resulting in Bb is ½ CC X Cc resulting in CC is ½ Therefore what is the probability that a cross between AABbCC X AaBbCc will produce offspring that are AaBbCC is ½ x ½ x ½ = 1/8
OK now practice this. 1. Parents = aaBbCC x aabbCc What is the probability that offspring will be aabbcc (recessive for all alleles)? 2. Parents are WwYYZz x wwyyZZ What is the probability that offspring will be WwYyZz? (heterozygous for all alleles)?
Q - When do we us the rule of addition? A - When we have two or more mutually exclusive events. Example: What is the probability that the offspring from a cross between WwYyZz x WwYYZZ will be WWYyZz or WWYYZZ? So first you do the same as above and use the rule of multiplication to work out the probability for each: WwYyZz x WwYYZZ WWYyZz = Ww ¼ x Yy ½ x Zz ½ = 1/16 WWYYZZ = WW ¼ x YY ½ x ZZ ½ = 1/16 Then you add the probabilities of these mutually exclusive events together. 1/16 + 1/16 = 2/16 or 1/8 so there is a 1/8 probability that the offspring from a cross between WwYyZz x WwYYZZ will be WWYyZz or WWYYZZ.
OK, now practice a few of these: 1. What is the probability that the offspring from a cross between SsTtVv x SsTtVV will be SSTTVV or ssttVV? 2. What is the probability that the offspring from a cross between CCDDEe x ccDdEe will be CcDDee or CCDdtEE? Slightly more complicated. 3. What is the probability that the offspring from a cross between CCDdee x CcDdEe will be dominant for at least 2 of the 3 characters? (hint: work out the probabilities for each offspring that has at least 2 dominant character using the multiplication rule then use the rule of addition) I will post the answers on Monday....