SOME SAMPLE PROBLEMS-EXAM 3

 

1.         Tay-Sachs is a recessive genetic disease in humans. If one parent is homozygous normal and the other is a carrier (heterozygous), what are the chances that their child will have this disease? NONE

 

2.         A woman who can roll her tongue (presumably dominant) is married to a man who cannot. Two of their four children can roll their tongues and two cannot. If A = roll tongue and a = cannot roll tongue, then what is the genotype of the parents? MALE=aa, FEMALE =Aa

 

3.         Some plants fail to produce chlorophyll, and this trait appears to be recessive. Many plants also self-pollinate. If we locate a pea plant that is heterozygous for this trait, self-pollinate it and harvest seeds, what are the likely phenotypes of these seeds when they germinate? 3:1

 

4.         In 1940, two researchers named Weiner and Landsteiner discovered that about 85 percent of the human population sampled possessed a blood cell protein labeled Rh positive (Rh+) and was found to be dominant over the absence of the blood factor (Rh-). Under normal Mendelian inheritance, which of the following statements is/are false?

A.        Two Rh+ parents could have an Rh- child.

B.         Two Rh- parents could have an Rh+ child.

C.        An Rh- child would require that both parents be carriers of at least one Rh- gene.

D.        It is possible with just one pair of parents to have children where some siblings are Rh- and some are Rh+.

E.         All of the above are false.

 

5.         If you had two guinea pigs of opposite sex, both homozygous, one black and one brown, but you didn't know which was the dominant characteristic, how would you find out the dominant color?

A.        Mate them and see what color the offspring are--that will be the dominant color.

B.         Mate them and see what color the offspring are--the other will be the dominant color.

C.        Mate them, then mate their offspring to see what color the next generation is--that will be the dominant color.

D.        Mate them together, then mate their offspring to see what color the next generation is--the other color will be the dominant color.

E.         None of the above are correct.

 

6.         How could you be certain that the guinea pigs used in the previous question are truly homozygous?

A.        The guinea pigs would be homozygous for black (or brown) coat color if each strain could be bred for many generations and only black (or brown) colored offspring were produced.

B.         If the immediate parents of the black (or brown) guinea pigs were both of that color, it proves they are homozygous.

C.        If a cross between the black and brown guinea pig produced four all black offspring, the black guinea pig would have to be homozygous for black coat color.

D.        Each of the above results would prove the black guinea pig was homozygous.

E.         Only microscopic examination of the guinea pig's genes could absolutely confirm homozygosity.

 

7.         In pea plants, the gene for round seed (R) is dominant, and wrinkled seeds (r) are recessive. The endosperm of the pea is also either starchy, a dominant gene (S), or waxy (s).What can be said of a fully heterozygous (or dihybrid) cross?

A.        It is impossible to secure offspring that are homozygous for both dominant genes.

B.         It is impossible to secure offspring that are homozygous for both recessive genes.

C.        It is impossible to secure offspring that are homozygous for one dominant gene such as round seed and homozygous recessive for the other recessive waxy gene.

D.        All of the above are impossible combinations in a dihybrid cross.

E.         All of the above are possible combinations in a dihybrid cross.

 

8.         Mr. X has been accused by Miss Z of fathering her child. The child's blood type is AB and Miss Z's blood type is A. Mr. X could NOT be the father if his blood type is__A or O______.

 

9.         Lethal genes are genes that result in the failure to develop a critical organ or metabolic pathway. They are nearly always recessive. Animal breeders who discover a unique trait and cross-breed to increase the occurrence of that trait often encounter a noticeable increase in lethal genes. Why?

A.        The lethal recessive gene may be linked to the desired trait gene.

B.              Spreading the gene among offspring of both sexes will increase the likelihood it will be sex-linked and expressed when alone.

C.        The cross-mating of closely related individuals, or inbreeding, increases chances two recessive genes will "meet" in offspring.

D.        Lethal genes may switch from recessive to dominant.

E.              Selecting for the desired trait may result in "uncovering" the lethal gene.

 

10.       A man with blood type AB could NOT be the father of a child with

A.        blood type A.

B.         blood type B.

C.        blood type AB.

D.        blood type O.

 

11.              Nondisjunction of the sex chromosome in a woman (assume normal meiosis in husband) could NOT lead to a child with the genotype

A.        XO.

B.         XXX.

C.        XXY.

D.        XYY.

 

12.       When Mendel established his mathematical model for pea plant inheritance, including his laws of segregation and independent assortment, he used a small number of traits that happened to be the same number as the numbers of chromosome pairs (yet to be discovered). What is the relationship between his traits and the number of chromosome numbers of peas?

A.        If he had selected two traits on the same chromosome, these genes would not have assorted independently of each other.

B.         Linkage groups would not cause him any trouble except for instances of crossing over.

C.        Linkage groups would not cause him any trouble except for cases of inversion.

D.        If all traits were on the same chromosome, they would be inherited in identical patterns without any variation.

E.         There would be no problem with linkage groups and he would have arrived at the same mathematical models of heredity particles.

 

13.       Animals generally build an organism using the diploid number of chromosomes.  However, insects in the order of ants, wasps and bees use a haploid-diploid system in which adults of one sex are formed with a haploid number of chromosomes.  This would mean that:

A.        a single set of chromosomes is sufficient to code for a functional individual.

B.         a female could determine the sex of the offspring by fertilizing or not fertilizing an egg.

C.        the males and females are not equally related to their mothers, considering the proportion of genes held in common.

D.        All of the above.

E.         This system in no way differs from regular diploid organisms.

 

14.       We presume that meiosis evolved later than mitosis.  What process(es) would have to evolve to make this transition?

A.              Homologous chromosomes would have to pair up.

B.         An additional division would have to occur but without a second DNA replication.

C.        Sister chromosomes do not separate at first anaphase; centromeres hold until second division.

D.        All of the above must evolve.

E.         All of the above occur in mitosis and none have changed in evolution of meiosis.

 

15.       In organisms that are parthenogenetic, females produce young that are female without any fertilization of eggs occurring.  Which process(es) could theoretically accomplish this through alteration of the normal meiotic cycle?

A.        A haploid egg develops into a female; females only complete meiosis I.

B.         A polar body fuses with the egg nucleus in place of the sperm and females complete meiosis.

C.        Meiosis stops before anaphase, telophase or cytokinesis occurs, leaving a full diploid individual.

D.        All of the above would theoretically produce parthenogenesis if the offspring were viable.

 

16. Polydactyly, or six fingers and/or six toes (dominant). Suppose a woman heterozygous for polydactyly mates with a normal man. What are the chances that their children will have six fingers and/or toes? 50%

 

17.Suppose a color blind man with type O blood mates with a woman with normal vision (homozygous) and type AB blood. What are the possible genotypes and phenotypes of their offspring? What are the probabilities of each? 50% A, 50% B, NO COLOR BLIND CHILDREN

 

18.A man has type A blood and his wife has type B. Their child has type O. What are the genotypes of the parents? AO, BO

 

 

 

 

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