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Practice Genetics Problems – Calculating Probabilities 1. The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd (AaBbCcDd x AaBbCcDd). Assuming independent assortment of these four genes, what are the probabilities that F2 offspring will have the following genotypes? A) aabbccdd B) AaBbCcDd C) AABBCCDD D) AaBBccDd E) AaBBCCdd 2. Karen and Steve each have a sibling with sickle-cell disease (recessive disorder). Neither Karen nor Steve nor any of their parents have the disease, and none of them have been tested to reveal sickle-cell trait. Based on this incomplete information, calculate the probability that if this couple has a child, the child will have sickle-cell disease. 3. Imagine that you are a genetic counselor, and a couple planning to start a family comes to you for information. Charles was married once before, and he and his first wife had a child with cystic fibrosis (a recessive disorder). The brother of his current wife, Elaine, died of cystic fibrosis. What is the probability that Charles and Elaine will have a baby with cystic fibrosis? (Neither Charles nor Elaine has cystic fibrosis.) Problem #1 Soultion: First, create Punnett Squares for the 4 individual Monohybrid crosses: A a B b A AA Aa B BB Bb a aA aa b bB bb C c D d C CC Cc D DD Dd c cC cc d dD dd Each Monohybrid cross can now have individual probabilities calculated for each genotype as follows: Monohybrid Cross #1, Aa x Aa AA = ¼; Aa = ½; aa = ¼ Monohybrid Cross #2, Bb x Bb BB = ¼; Bb = ½; bb = ¼ Monohybrid Cross #3, Cc x Cc CC = ¼; Cc = ½; cc = ¼ Monohybrid Cross #4, Dd x Dd DD = ¼; Dd = ½; dd = ¼ Part A) now wants us to consider the probability of having an offspring with the genotype of “aabbccdd,” so we must consider the four individual probabilities linked (“aa” and “bb” and “cc” and “dd”), leading us to multiplication of the individual probabilities as follows: aa x bb x cc x dd = ? ¼ x ¼ x ¼ x ¼ = 1/256 Part B) now wants us to consider the probability of having an offspring with the genotype of “AaBbCcDd,” so we must consider the four individual probabilities linked (“Aa” and “Bb” and “Cc” and “Dd”), leading us to multiplication of the individual probabilities as follows: Aa x Bb x Cc x Dd = ? ½ x ½ x ½ x ½ = 1/16 Part C) now wants us to consider the probability of having an offspring with the genotype of “AABBCCDD,” so we must consider the four individual probabilities linked (“AA” and “BB” and “CC” and “DD”), leading us to multiplication of the individual probabilities as follows: AA x BB x CC x DD = ? ¼ x ¼ x ¼ x ¼ = 1/256 Part D) now wants us to consider the probability of having an offspring with the genotype of “AaBBccDd,” so we must consider the four individual probabilities linked (“Aa” and “BB” and “cc” and “Dd”), leading us to multiplication of the individual probabilities as follows: Aa x BB x cc x Dd = ? ½ x ¼ x ¼ x ½ = 1/64 Part E) now wants us to consider the probability of having an offspring with the genotype of “AaBBCCdd,” so we must consider the four individual probabilities linked (“Aa” and “BB” and “CC” and “dd”), leading us to multiplication of the individual probabilities as follows: Aa x BB x CC x dd = ? ½ x ¼ x ¼ x ¼ = 1/128 Problem #2 Solution: First, create a list to summarize the information provided above: Karen – No Disease Karen’s Sibling – Disease Karen’s Parents – Neither has Disease Steve – No Disease Steve’s Sibling – Disease Steve’s Parents – Neither has Disease Second, reason through what each individual above has for a genotype based on the phenotypes provided – we will assign a “S” for the dominant allele and a “s” for the recessive allele: Karen = SS or Ss Steve = SS or Ss Karen’s Sibling = ss Steve’s Sibling = ss Karen’s Parents = both are Ss Steve’s Parents = both are Ss We know that Karen’s sibling and Steve’s sibling must be “ss” because each sibling has the disease (this only manifests when there are 2 copies of the “s” allele). Where did each sibling get those alleles? – from their parents. Therefore, Karen’s and Steve’s parents must have at least one copy of the “s” allele; however, the parents do not have the disease, therefore they must be heterozygotes (carriers with genotype of “Ss”). Still we do not know what Karen’s or Steve’s genotype is specifically – it could be either “SS” or “Ss” since either of these is a possibility that yields no disease, resulting from having heterozygote parents (Ss x Ss) – we know for sure that neither Karen nor Steve can be “ss” because they do not have the disease. See below: Karen’s Parental Cross S s Possible genotypes S SS Ss for Karen s sS ss Not a possible genotype for Karen Steve’s Parental Cross S s Possible S SS Ss genotypes for Steve s sS ss Not a possible genotype for Steve Third, we now need to know the probability of Karen and Steve each being heterozygotes (“Ss”) since this is the only way there is the possibility of having a child with Sickle-cell Disease. As we can see from above, Karen has a 2 in 3 chance of being “Ss” and so does Steve. Therefore the probability that both are heterozygous (Karen = Ss and Steve = Ss), we take 2/3 x 2/3 = 4/9. Fourth, we must now calculate the final individual probability of Karen and Steve’s child having the disease: Karen and Steve Cross S s S SS Ss s sS ss Child with disease = ¼ probability Lastly, we put all of this together in determining the combined probability that Karen is heterozygous and Steve is heterozygous and that their child has the disease 2/3 x 2/3 x ¼ = 4/36 = 1/9 Problem #3 Solution: Use problem #2 as a guide to reason your way through to the realization that the probability of Charles being heterozygous is 1, and the probability of Elaine being heterozygous is 2/3, and the probability that their child has the disease is ¼ therefore, 1 x 2/3 x ¼ = 2/12 = 1/6