I do not understand how to do this. We can make up the the number for the individuals but I do not understand how to get the phenotype and genotype ratios and the other parts of the questions. Please help me figure this out...Questions are numbered in red. THANKS
Carefully count the number of individuals in any group (about 30), that have bent little finger and attached ear lobes. Count the number of individuals that have attached ear lobes but straight little fingers. Count the number of individuals that do not have attached ear lobes but bent little fingers. Finally, count the number of individuals that do not have attached ear lobes and have straight little fingers.
Attached ear lobes and bent little finger:
Attached ear lobes and straight little finger:
Non-attached ear lobes and bent little finger:
Non-attached ear lobes and straight little finger:
These numbers that you have just counted represent the phenotypic ratio you achieve in a dihybrid cross between individuals heterozygous in both traits. These numbers can also be used in support of Mendel’s Law of Independent Assortment, which states that when allelic genes are separated into gametes, this is done so, independently of the allelic genes on a different pair of chromosomes of the same organism.
In order to demonstrate this, count the number of AA individuals
(7) ; Aa individuals
(8) ; aa individuals
(9) . The genotypic ratio is
(10) : : . The phenotypic ratio would be
(11) : .
Count the number of BB individuals
(12) : Bb individuals
(13) ; bb
(14) individuals Once again, the genotypic ratios and the phenotypic ratios are the same as if these traits had been observed individually. This shows us that the inheritance pattern of each trait is independent of the influence of the other traits under consideration. Because of this, the probability laws of independent events may be applied. This law states that the chance of 2 separate events taking place at the same time is the product of the chance that each event will take place by itself. Let’s use the above situation as an example. We see that there is a 50% chance of getting the A allele in the gamete. There is also a 50% chance that the A allele will be in a gamete. The chance that the B allele will end up in the same gamete as the A allele then is the product (multiply) of the chance for each allele ending up in a gamete or 50% x 50% = 25%. If you are using fractions then it would simply be 1/2 x 1/2 = 1/4.
Since we can easily look at many traits at one time using this method let’s look at 4 independent gene pairs at once.
Aa, Bb, Cc, Dd,
What would be the chance (probability) of producing a gamete with the genes abcd? I'll start for you, getting a from Aa is 1/2, getting b from Bb is 1/2, etc., etc.
½ x ½ x ½ x ½ = 1/8
For the independent gene pairs, Aa, bb, CC, Dd, and ee, what would be the chance of producing a gamete with the genes AbCde? CC probability is 1.0
(15) For the same independent gene pairs as above, what would be the chance of producing a gamete with the genes ABCDE?
(16)
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