Red-green Color Blindness and Hemophilia A are both sex-linked traits in humans. Both genes are on the X-chromosome and are tightly linked.
Consider the pedigree below. Individual 1 is homozygous normal. Individual 2 is both red-green color blind and a hemophiliac. Individual 6 is the as yet unborn son of 4 and 5.
What is the probability that their son, 6, will be both color blind and a hemophiliac? Express your answer as a number between 0.0 and 1.0. You may use as many significant figures as necessary.
You tell about individuals 1 and 2, but not 4 and 5. Who are 4 and 5?
For 1 and 2, is the person with the recessive colorblindness and hemophilia male or female? If male, none of sons will be either, since they only get the Y-chromosome from him. If female, all the sons will have both, since she has to be homozygous for the recessive genes.
0.5
Traits like human height and skin color shows a wide range of phenotypes, as opposed to just two ot three. From the standpoints of genese and alleles, explain why is this so.
To determine the probability that their son, Individual 6, will be both color blind and a hemophiliac, we need to understand the inheritance patterns of these sex-linked traits.
In humans, the X chromosome is important for the inheritance of these traits. Males have one X chromosome and one Y chromosome (XY), while females have two X chromosomes (XX). Since the genes for red-green color blindness and hemophilia A are tightly linked on the X chromosome, they are inherited together more often than not.
Now let's analyze the given pedigree to determine the probability.
Individual 1 is homozygous normal, which means they have two normal X chromosomes (XX). Since both genes for color blindness and hemophilia are recessive, Individual 1 does not carry the genes for these traits.
Individual 2 is both color blind and a hemophiliac. This means that Individual 2 must have inherited a single X chromosome from their mother (who is not color blind or a hemophiliac), carrying the recessive alleles for both traits.
Individual 4 is a carrier for both traits, meaning they have one normal X chromosome and one X chromosome with the recessive alleles for color blindness and hemophilia.
Individual 5 is also a carrier, with the same genotype as Individual 4.
Now, let's consider the probabilities for Individual 6, the unborn son of 4 and 5. Since Individual 4 carries the genes for both traits and Individual 5 is also a carrier, they each have a 50% chance of passing on the recessive alleles for color blindness and hemophilia to their child.
To determine the probability, we need to multiply the chances of passing on both traits together. Since the inheritance of these traits is independent of each other, we can multiply the probabilities.
The probability of passing on the recessive allele for color blindness is 1/2 (50%), and the probability of passing on the recessive allele for hemophilia is also 1/2 (50%).
Therefore, the probability of Individual 6 being both color blind and a hemophiliac is (1/2) * (1/2) = 1/4 = 0.25.
So, the probability that their son, Individual 6, will be both color blind and a hemophiliac is 0.25 or 25%.