Genetics Questions

In cats with the Manx trait, the M (dominant) allele causes a short or absent tail, whereas the m allele confers a normal, long tail. Cats of genotype MM die as embryos. If two Manx cats mate, what is the probability that each living kitten has a long tail?

1/3

An organism with the genotype AaBbCCdd is crossed with one that is AAbbccDd. What is the probability of having an offspring with the genotype AabbCcDd?

1/8 (½ x ½ x 1 x½ )

A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the normal number of digits. Extra digits is a dominant trait. What fraction of this couple's children would be expected to have extra digits? What is the probability that each of the following pairs of parents will produce the indicated offspring (assume independent assortment of all gene pairs)? 1. AABBCC x aabbcc —— AaBbCc 2. AABBCc x AaBbCc —— AAbbCC 3. AaBbCc x AaBbCc —— AaBbCc 4. aaBbCC x AABbcc —— AaBbCc

1= 1; 2= 1/32 (2 x 4 x ); 3 = 1/8 (½ x ½x %); 4 = ½ (1x ½x 1)

In mice, the spinning behavior is caused by a dominant gene that affects the mouse's equilibrium. This gene is lethal if two alleles are present. Two "spinning mice" are mated. What are the phenotypes of the offspring and in what proportion?

2: 1 spinning : normal

If the dominant allele K is necessary for hearing, and the dominant allele M of another independent locus (i.e., not linked) results in deafness no matter what other genes are present, what percentage of the offspring produced by the cross kkMm x Kkmm will be deaf?

3/4

Color in a species of duck is determined by one gene with three alleles. Ducks homozygous for allele B are blue, ducks homozygous for allele Y are yellow and ducks homozygous for allele i are white. Alleles B and Y are dominant over allele i, but BY heterozygòtes are green. How many genotypes are possible, and how many phenotypes are possible in a flock of ducks that contains all the possible combinations of these three alleles?

6 genotypes ( YY, Yi, BB, Bi, YB, ii) , 4 phenotypes (Yellow, Blue, Green, White)

assume the following pigment-producing pathway in a plant: Also assume that the A and B alleles produce functional enzyme, while the a and b alleles produce no functional enzyme. Assume that one functional copy of an enzyme is sufficient to catalyze the reaction. Predict the colors of the progeny if you cross AaBb with AaBb

9 purple: 4 colorless: 3 red

A black guinea pig crossed with an albino guinea pig produced 12 black offspring. When the albino was crossed with a second black one, 7 blacks and 5 albinos were obtained. What is the best explanation for this genetic situation? Write genotypes for the parents, gametes, and offspring.

Black (B)is dominant over albino (b). First male is BB, second male is Bb, mother is bb. black F1 are Bb, albino F1 are bb.

Some children are born with recessive traits (and, therefore, must be homozygous for the recessive allele specifying the trait), even though neither parent exhibits the trait. What can account for this?

Both parents carry at least one of the recessive genes, even if only in the heterozygous condition (in which the trait would not be expressed in the parent). Since it is recessive, the trait is not manifest until they produce an offspring who is homozygous.

Many animals and plants bear recessive alleles for albinism, a condition in which homozygous individuals lack certain pigments. An albino plant, for example, lacks chlorophyll and is white, and an albino human lacks melanin. If two normally pigmented persons heterozygous for the same albinism allele marry, what proportion of their children would you expect to be albino?

Albinism, a, is a recessive gene. If heterozygotes mated you would have the following: Clearly one-fourth would be expected to be albinos.

The illustration describes Mendel's cross of wrinkled and round seed characters. What is wrong with this diagram? (Hint: Do you expect all the seeds in a pod to be the same?)

Alleles segregate in meiosis, and the products of that segregation are contained within a pod. Each pea is a zygote. In this diagram, the segregation is incorrectly shown as being between pods, each pod shown as uniformly wrinkled or round.

Your instructor presents you with a Drosophila with red eyes, as well as a stock of white-eyed flies and another stock of flies homozygous for the red-eye allele. You know that the presence of white eyes in Drosophila is caused by homozygosity for a recessive allele. How would you determine whether the single red-eyed fly was heterozygous for the white-eye allele?

Breed the fly to one from the white-eyed stock. If half of the offspring are white-eyed, then your fly is a heterozygote.

Total color blindness is a rare hereditary disorder among humans. Affected individuals can see no colors, only shades of gray. It occurs in individuals homozygous for a recessive allele, and it is not sex- linked. A man whose father is totally color blind intends to marry a woman whose mother is totally color blind. What are the chances they will produce offspring who are totally color blind?

It expresses dominant inheritance.

A couple both work in an atomic energy plant, and both are exposed daily to low-level background radiation. After several years, they have a child who has Duchenne muscular dystrophy, a recessive genetic defect caused by a mutation on the X chromosome. Neither the parents nor the grandparents have the disease. The couple sue the plant, claiming that the abnormality in their child is the direct result of radiation-induced mutation of their gametes, and that the company should have protected them from this radiation. Before reaching a decision, the judge hearing the case insists on knowing the sex of the child. Which sex would be more likely to result in an award of damages, and why?

It would be very difficult to establish negligence, especially if the child is a boy. If the child's maternal grandmother was a carrier, there's a 50:50 chance that the mother is a carrier. If the mother is a carrier, there's a 50:50 chance that a boy will express the disease. One would be suspicious about the causes of Duchenne muscular dystrophy in a girl. In the first place,

An inherited trait among humans in Norway causes affected individuals to have very wavy hair, not unlike that of a sheep. The trait, called woolly, is very evident when it occurs in families; no child possesses woolly hair unless at least one parent does. Imagine you are a Norwegian judge, and you have before you a woolly-haired man suing his nomal-haired wife for divorce because their first child has woolly hair but their second child has normal hair. The husband claims this constitutes evidence of his wife's infidelity. Do you accept his claim? Justify your decision.

It would not be possible on the basis of the information presented to substantiate a claim of infidelity. You do not know if the woolly trait is the result of a single gene product, or even if the trait is dominant or recessive. Assuming for the moment that it was the effect of a single dominant allele W, the man would still be a heterozygote for the gene and, when mated to a recessive homozygous female, would expect to produce woolly-headed offspring only one-half the time.

A normally pigmented man marries an albino woman. They have three children, one of whom is an albino. What is the genotype of the father?

Let a-albino. The genotype of the father is Aa.

A couple with a newborn baby is troubled that the child does not resemble either of them. Suspecting that a mix-up occurred at the hospital, they check the blood type of the infant. It is type O. As the father is type A and the mother type B, they conclude a mix-up must have occurred. Are they correct?

No. IAIO x IBIO —> IAIB or IAIO or IBIO or IOIO

In human beings, Down syndrome, a serious developmental abnormality, results from the presence of three copies of chromosome 21 rather than the usual two copies. If a female exhibiting Down syndrome mates with a normal male, what proportion of her offspring would you expect to be affected?

One-half of her offspring would be expected to be affected.

A rooster with gray feathers is mated with a hen of the same phenotype. Among their offspring, 15 chicks are gray, 6 are black, and 8 are white. What is the simplest explanation for the inheritance of these colors in chickens? What offspring would you predict from the mating of a gray rooster and a black hen?

Simplest explanation is incomplete dominance: Black = CBCB, Gray = CBCW, White = CWCW Therefore predict gray x black = 1:1 gray: black

Among Hereford cattle there is a dominant allele called polled; the individuals that have this allele lack horns. Suppose you acquire a herd consisting entirely of polled cattle, and you carefully determine that no cow in the herd has horns. Some of the calves born that year, however, grow horns. You remove them from the herd and make certain that no horned adult has gotten into your pasture. Despite your efforts, more horned calves are born the next year. What is the reason for the appearance of the horned calves? If your goal is to maintain a herd consisting entirely of polled cattle, what should you do?

Somewhere in your herd you have cows and bulls that are not homozygous for the dominant gene "polled." Since you have many cows and probably only one or some small number of bulls, it would make sense to concentrate on the bulls. If you have only homozygous "polled" bulls, you could never produce a horned offspring regardless of the genotype of the mother. The most expedient thing to do would be to keep track of the matings and the phenotype of the offspring resulting from these matings and render ineffective any bull found to produce horned offspring.

In 1986, National Geographic magazine conducted a survey of its readers' abilities to detect odors. About 7% of Caucasians in the United States could not smell the odor of musk. If neither parent could smell musk, none of their children were able to smell it. On the other hand, if the two parents could smell musk, their children generally could smell it, too, but a few of the children in those families were unable to smell it. Assuming that a single pair of alleles governs this trait, is the ability to smell musk best explained as an example of dominant or recessive inheritance?

The ability to smell musk is an example of dominant inheritance.

You inherit a racehorse and decide to put him out to stud. In looking over the stud book, however, you discover that the horse's grandfather exhibited a rare disorder that causes brittle bones. The disorder is hereditary and results from homozygosity for a recessive allele. If your horse is heterozygous for the allele, it will not be possible to use him for stud, since the genetic defect may be passed on. How would you determine wherher your horse carries this allele?

The best thing to do would be to mate the racehorse to several mares homozygous for the recessive gene that causes the brittle bones. Half of the offspring would be expected to have brittle bones if the racehorse were a heterozygous carrier of the disease gene. Although you could never be 100% certain your horse was not a carrier, you could reduce the probability to a reasonable level.

A normally pigmented man marries an albino woman. They have three children, one of whom is an albino. What is the genotype of the father?

The chances are 25%.

A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the normal number of digits. Extra digits is a dominant trait. What fraction of this couple's children would be expected to have extra digits?

The man must be heterozygous, so half of this couple's children should have extra digits.

The annual plant Haplopappus gracilis has 2 pairs of chromosomes 1 and 2. In this species, the probability that two traits a and b selected at random will be on the same chromosome is equal to the probability that they will both be on chromosome 1 (0.5 x 0.5 = 0.25), plus the probability that they will both be on chromosome 2 (0.5 x 0.5 = 0.25), for an overall probability of 0.5. In general, the probability that two randomly selected traits will be on the same chromosome is equal to 1/n where n is the number of chromosome pairs. Humans have 23 pairs of chromosomes. What is the probability that any two human traits selected at random will be on the same chromosome?

The probability of getting two genes on the same chromosome is (1/23), but since there are 23 chromosomes the probability that they are on the same chromosome is (1/23)^2 x 23, or 1/23.

A woman is married for the second time. Her first husband has blood type A and her child by that marriage bas type O. Her new husband has type B blood, and when they have a child its blood type is AB. What is the woman's blood genotype and blood type?

The woman's blood genotype and blood type is AO.

In 1981, a stray black cat with unusual rounded, curled black ears was adopted by a family in California. Hundreds of descendants of the cat have since been born, and cat fanciers hope to develop the "curl" cat into a show breed. Suppose you owned the first curl cat and wanted to develop a true-breeding variety. How would you determine whether the curl allele is dominant or recessive? How would you select for true-breedung cats? How would you know they are true-breeding?

To determine if curl is dominant or recessive cross with normal cats. Assume that the allele is rare, so if dominant half of the kittens will have curled ears, half will have normal ears. If recessive, all of the kittens will have normal ears. To get true-breeding recessives, cross the F1 with each other, and hopefully you will get some curled ears in the F2. Cross them with each other and you will have a true-breeding line. If it is dominant you will need to cross curled-eared F1 with each other. They will all be heterozygous, but 1/3 of the F2 will be homozygous curled. To find out, cross them with normal ears, and retain the cats that give all curled ears in this test- cross and cross them with each other.

You collect two individuals of Drosophila, one a young male and the other a young, unmated female. Both are normal in appearance, with the red eyes typical of Drosophila. You keep the two flies in the same bottle, where they mate. Two weeks later, the offspring they have produced all have red eyes. From among the offspring, you select 100 individuals, some male and some female. You cross each individually with a fly you know to be homozygous for the recessive allele sepia, which produces black eyes when homozygous. Examining the results of your 100 crosses, you observe that in about half of the crosses, only red-eyed flies were produced. In the other half, however, the progeny of each cross consists of about 50% red-eyed flies and 50% black-eyed flies. What were the genotypes of your original two flies?

To solve this problem, first look at the second cross, where the individuals were crossed with the homozygous recessive sepia flies se/se. In one case, all the flies were red-eyed: Unknown genotype The only way to have all red-eyed flies when bred to homozygous sepia flies is to mate the sepia fly with a homozygous red-eyed fly. In the other case, half of the offspring were black-eyed and the other half red-eyed: Unknown genotype The unknown genotype in this case must have been Se/se, since this is the only mating that will produce the proper ratio of sepia-eyed flies to red-eyed flies. Since the ratio of this unknown genotype and one previously determined was 1:1, we must deduce the genotype of the original flies, which when mated, will produce a 1:1 ratio of Se/se to Se/Se flies. Unknown Original 1 You can see from this diagram that if one of the original flies was homozygous for red yes and the other was a heterozygous individual, the proper ratio of heterozygous and homozygous offspring would be obtained.

How many chromosomes would you expect to find in the karyotype of a person with Turner syndrome?

You could expect 45 chromosomes (44 autosomes plus one extra copy of an X).

In Drosophila, the allele for dumpy wings (d) is recessive to the normal long-wing allele (d+), and the allele for white eye (w) is recessive to the normal red-eye allele (w+). In a cross of d+d+w+w x d+dww, what proportion of the offspring are expected to be "normal" (long wings, red eyes)? What proportion are expected to have dumpy wings and white eyes?

Your mating of d d Ww and d dww individuals would look like the following: Long-wing, red-eyed individuals would result from eight of the possible 16 combinations, and dumpy, white-eyed individuals would never be produced.

Hemophilia is a recessive sex-linked human blood disease that leads to failure of blood to clot normally. One form of hemophilia has been traced to the royal family of England, from which it spread throughout the royal families of Europe. For the purposes of this problem, assume that it originated as a mutation either in Prince Albert or in his wife, Queen Victoria. a. Prince Albert did not have hemophilia. If the disease is a sex-linked recessive abnormality, how could it have originated in Prince Albert, a male, who would have been expected to exhibit sex- linked recessive traits? b. Alexis, the son of Czar Nicholas II of Russia and Empress Alexandra (a granddaughter of Victoria), had hemophilia, but their daughter Anastasia did not. Anastasia died, a victim of the Russian revolution, before she had any children. Can we assume that Anastasia would have been a carrier of the disease? Would your answer be different if the disease had originated in Nicholas II or in Alexandra?

a. It could have originated as a mutation in his germ cell line. b. Since their son Alexis was a hemophiliac, the disease almost certainly originated with Alexandra; Nicholas II would have contributed only a silent Y-chromosome to Alexis's genome. There is a 50% chance that Anastasia was a carrier.

Mabel's sister died of cystic fibrosis as a child. Mabel does not have the disease, and neither do her parents. Mabel is pregnant with her first child. If you were a genetic counselor, what would you tell her about the probability that her child will have cystic fibrosis?

c is a rare allele, and Mabel's child can develop cystic fibrosis only if her husband also carries allele: [probability that husband is a carrier] = 1/21, therefore probability that any one of his gametes has the c allele is 1/42 [probability that Mabel is a carrier] = 2/3, because we know that her parents are carriers but she doesn't suffer. Therefore, probability that any given egg has the c allele is 1/3 probability of having a child with cystic fibrosis is 1/42 * 1/3 = 1/126