In humans, brown eyes (B) are dominant over blue (b)*. A brown-eyed man marries a blue-eyed woman and they have three children, two of whom are brown-eyed and one of whom is blue-eyed. Draw the Punnett square that illustrates this marriage. What is the man’s genotype? What are the genotypes of the children?
In dogs, there is an hereditary deafness caused by a recessive gene, “d.” A kennel owner has a male dog that she wants to use for breeding purposes if possible. The dog can hear, so the owner knows his genotype is either DD or Dd. If the dog’s genotype is Dd, the owner does not wish to use him for breeding so that the deafness gene will not be passed on. This can be tested by breeding the dog to a deaf female (dd). Draw the Punnett squares to illustrate these two possible crosses. In each case, what percentage/how many of the offspring would be expected to be hearing? deaf? How could you tell the genotype of this male dog? Also, using Punnett square(s), show how two hearing dogs could produce deaf offspring.
In humans, there is a gene that controls formation (or lack thereof) of muscles in the tongue that allow people with those muscles to roll their tongues, while people who lack those muscles cannot roll their tongues. The ability to roll one’s tongue is dominant over non-rolling. The ability to taste certain substances is also genetically controlled. For example, there is a substance called phenylthiocarbamate (PTC for short), which some people can taste (the dominant trait), while others cannot (the recessive trait). The biological supply companies actually sell a special kind of tissue paper impregnated with PTC so students studying genetics can try tasting it to see if they are tasters or non-tasters. To people who are tasters, the paper tastes very bitter, but to non-tasters, it just tastes like paper. Let’s let R represent tongue-rolling, r represent a non-roller, T represent ability to taste PTC, and t represent non-tasting.
Suppose a woman who is both a homozygous tongue-roller and a non-PTC-taster marries a man who is a heterozygous tongue-roller and is a PTC taster, and they have three children: a homozygous tongue-roller who is also a PTC taster, a heterozygous tongue-roller who is also a taster, and a heterozygous tongue-roller who is a non-taster. If these parents would have a bunch more children so that they had 12 in all, how many of those 12 would you expect to be non-tasters who are homozygous for tongue-rolling? If the first child (the homozygous tongue-roller who is also a PTC taster) marries someone who is heterozygous for both traits, draw the Punnett square that predicts what their children will be.
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If the man is both Rr and Tt (How do we know that?), he would be RrTt and so could produce gametes with either R or r and either T or t (one allele for each gene). There are two choices for the first trait (R or r). No matter which of those go into a given sperm, there are still two choices for the second trait (T or t). Therefore, a total of 2 × 2 = 4 possible types of gametes (sperm) may be produced. What kind(s) of gametes (sperm) can he produce?
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4) There is another gene that codes for another, different antigen that also occurs on the surface of our RBCs, and technically, that gene also has multiple alleles. However, most people either have or do not have one particular allele called the “d” allele. This gene codes for an antigen that is called “Rh factor” because it was first discovered in Rhesus monkeys. People who have instructions to “make d antigen” are referred to as Rh+ (the allele is often symbolized by the letter “R”), while those who have “I don’t know how to make d antigen” instructions are called Rh– (the allele can be symbolized by “r”). Since this is a totally separate gene than the ABO blood group, if you’re doing a genetic cross that involves both ABO and Rh, that would be a dihybrid cross.
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Ms. Branch, Ms. Davis, and Ms. Rucker all entered the same hospital and gave birth to baby girls on the same day, and all three babies were taken to the nursery to receive care, there. Someone later claimed that the hospital mixed up the babies. As a hospital administrator, it is your job to make sure that each pair of parents has the correct baby, so you order blood typing to be done on all the parents and all the babies. Here are the results:
In Guinea pigs, black hair (B) is dominant over white (b), rough coat texture (R) is dominant over smooth (r), and short hair (S) is dominant over long hair (s). Assuming these genes are on separate chromosomes, draw the Punnett square for a cross between a homozygous black, rough, short-haired Guinea pig and a white, smooth, long-haired one. What would the phenotype(s) of the offspring be? If two of the F1offspring were crossed, draw the Punnett square for this cross. Hint: first make a list of the possible gametes, making sure each has exactly one copy of each of the genes (one allele for each gene). What would the genotype and phenotype ratios be for the F2 generation?