Biology 198
PRINCIPLES OF BIOLOGY
Mendelian Genetics problems
Updated: 21 August 2000
Gregor Mendel, an Austrian monk, revealed through numerous experiments with pea plants that offspring are simply not "blends" of their parents. Rather, he clearly demonstrated that traits tend be passed to offspring in a "particulate" fashion. Indeed, if the blending theory were true, then everyone would eventually look about the same. Who knows; perhaps the most favorable phenotypes would be similar to Rodney Dangerfield and Phyllis Diller.
Mendel decided to perform some crosses with his plants to test the blending theory. First, he crossed tall plants (i.e. long stemmed) which had parents and grandparents that had all been tall, with short plants (short stemmed) which had parents and grandparents that had all been short. The plants he crossed were termed the P (parental) generation. He found that the offspring, or F1 (filial) generation, were 100% tall. However, when he crossed these F1 plants together he found that the F2 generation represented a mixture: 3/4 were tall, and 1/4 were short. Whenever he performed such crosses, he always came out with these mathematical ratios. These mathematical ratios are very important because changes in ratios can sometimes be indicators that evolution is occurring.
How can this "particulate" nature of heredity be explained? Well, Mendel had a background in mathematics and through years of crossing and eating peas he came up with the following hypothesis: that there must be 2 factors (now called alleles) for each trait (now called genes), and that these factors (alleles) behave as distinct "particles" when passed to the offspring. Some of these traits are dominant (i.e. when present they are expressed), whereas others are recessive and their expression can be "masked" by dominant alleles. Offspring receive one of the factors (an allele) from one parent, and receive the other factor from the other parent.
Lets assign letters to each trait and perform some genetic crosses using a single gene. This is termed a monohybrid cross (simply a cross between two individuals where we look at a single gene). In pea plants, tall is dominant and is represented by a big T. Any time a big T is present, the plant is tall. When no big T is present, the plant will appear short. Since Mendel's tall parent plants were all true breeding (i.e. only produced tall plants) both of their factors must have been big so they are represented as "TT." If any small factors were present, they would have certainly have shown up at some time during his observations of the parent and grandparent plants. This "TT" is the genotype (i.e., the genetic makeup representing the 2 alleles that are present). Because the short plants were true breeding (i.e. all offspring from crosses yielded 100% short plants), their genotype must have been "tt." We can visually show such a cross between two parents by using a Punnett's square:
| t | t | |
|---|---|---|
| T | Tt | Tt |
| T | Tt | Tt |
Note that along the top we write in the two alleles that segregated into the different gametes from one parent prior to mating (light blue color). Along the left side, we write in the two alleles from the other parent (pale green color). The resulting diploid organisms (termed F1s) within the squares themselves represent the offspring. As one can see, 100% of all offspring are heterozygous (Tt). Since each organism possesses at least one big T, and big T is dominant to little t, whatever effects may be produced by the little t are totally masked. All offspring in this cross express the dominant characteristic for tallness. Thus, the genotype of all offspring is Tt, and the phenotype (what you see) is tall. In reality, most combinations of alleles in living organisms do not display complete dominance or recessiveness. Rather, both usually contribute to the outcome of the offspring and most characteristics are under the control of multiple genes. However, it is likely that Mendel spent considerable time observing his pea plants prior to his experiments and when the time came to get down and dirty with his experiments he probably chose only those characteristics that seemed to make sense to him.
Now then, lets perform Mendel's next experiments where he crossed the F1s to each other. You'll note from the above that all offspring were heterozygous (Tt). In this type of cross, he found that 3/4 of the F2 generation were tall, but that the recessive characteristic reappeared in 1/4 of the offspring. The cross can be visualized as follows:
| T | t | |
|---|---|---|
| T | TT | Tt |
| t | Tt | tt |
Indeed, we can now see why 1/4 of the individuals were short since only one in four (lower right) possess the tt genotype; i.e. no big T is present in 25% of the offspring. Thus, the genotype of the F2 population is 25% TT, 50% Tt, and 25% tt (a ratio of 1:2:1). The phenotype is different, however. Since 3/4 possess at least one big T, they will be tall. So, the phenotypic ratio is 3:1 (tall vs short).
The totally recessive individuals are highly useful in genetics. Whenever you see one, you automatically know the entire genotype (i.e. both alleles) for that gene. For instance, if you have a tall individual, you know that at least one big T is present but you don't know if the second allele is "T" or "t." Not so for the short pea plants as they can only be "tt." These recessive offspring are extremely valuable in genetics. You not only know the genotype of the individual in question, but you also know that each of the parents carries at least one recessive allele whether you can see it or not (i.e., the offspring HAD to get a little "t" from EACH parent). These recessive individuals can be used to "test" an unknown plant and probe for any hidden, recessive alleles. This is called a testcross.
The results of Mendel's crosses allowed him to formulate his Law of Segregation, which states that each organism contains two factors (i.e. alleles) for each trait, and the factors segregate during the formation of gametes so that each gamete contains only one factor for each trait. What this means is that the alleles of an organism exist as pairs (we now know on separate, homologous chromosomes) and that one member of the pair enter different gametes (i.e., we now know that the homologous chromosomes separate during meiosis I).
OK. You now, hopefully, understand some of the basics of how alleles segregate into the gametes and how the gametes fuse to form a zygote. Using what we've learned above, lets see if you can work the following two monohybrid genetics problems.
PROBLEM 1.
Hypothetically, brown color (B) in naked mole rats is dominant to white color (b). Suppose you ran across a brown, male, naked mole rat in class and decided to find out if he was BB or Bb by using a testcross. You'd mate him to a white (totally recessive) female, and examine the offspring produced. Now, if only 2-3 offspring were born and they were all brown, you'd still be uncertain whether he was BB or Bb (for instance, even though the odds are 50:50 that you will produce a boy or girl, there are plenty of people that produce 4-5 girls and never a boy and vice versa). But, if the mole rats produce 50 offspring and all are brown, then it is likely that no hidden alleles are present and that the male is BB. But, what if white offspring are produced? You'd know that the brown parent had a hidden little "b" allele. So, what you need to do is perform a testcross on this brown, male, heterozygous, naked mole rat. What are the expected genotypic and phenotypic ratios of such a cross?
PROBLEM 2.
What if you bred some snap dragons and crossed a homozygous red plant (RR) with a homozygous white plant (rr)? In botony, "true breeding" means homozygous. In this case, 100% of the F1 individuals would be pink! This is an example of "incomplete dominance," where both alleles contribute to the outcome. In some cases of incomplete dominance, both alleles might contribute equally so one allele would produce red pigment and the other white; thus, a pink plant appears. In another case, one allele may be non-functional. Although in many cases only a single allele is needed, perhaps in this case only one-half the amount of needed pigment is produced and so pink is due the low amount of red pigment in the petals. Who knows. Anyway, use a Punnett's square and set up a cross between a homozygous red plant and a homozygous white plant. Then, take the resulting offspring and cross these among themselves as well (i.e. F1 x F1). Then, determine the phenotypic and genotypic ratios.
The above examples involved a single gene only, and a single set of alleles. But, what if TWO different genes are involved. Well, that makes the crosses more complicated and we term this a dihybrid cross. As long as the genes are on separate chromosomes, things go pretty easily. Only the Punnett squares are larger (16 squares rather than 4). So, lets cross an individual who is homozygous dominant for gene A (AA) and homozygous dominant for gene B (BB) with someone who is homozygous recessive for gene A (aa) and homozygous recessive for gene B (bb). If you follow the movement of alleles into the gametes, for instance with the AABB, the alleles will assort as follows: the first A in the AABB goes with the first B (I've underlined this set) and is placed in column 1 (light blue); in the second pairing the first A goes with the second B (column 2); in the third pairing the second A goes with the first B (column 3); and in the fourth pairing the second A goes with the second B (column 4). Then, do the same for the aabb individual and place these pairs of alleles in the four rows (light green). As you will note below, all offspring are AaBb and will possess the same phenotype.
| AB | AB | AB | AB | |
|---|---|---|---|---|
| ab | AaBb | AaBb | AaBb | AaBb |
| ab | AaBb | AaBb | AaBb | AaBb |
| ab | AaBb | AaBb | AaBb | AaBb |
| ab | AaBb | AaBb | AaBb | AaBb |
As you will note from the above, the genotypes of all individuals is AaBb, and the phenotype (whatever it is) will be all the same. Now then, what if we crossed two of the F1 generation (heterozygotes)? The resulting Punnett's square should contain 9 different genotypes and 4 different phenotypes, as follows:
| AB | Ab | aB | ab | |
|---|---|---|---|---|
| AB | AABB | AABb | AaBB | AaBb |
| Ab | AABb | AAbb | AaBb | Aabb |
| aB | AaBB | AaBb | aaBB | aaBb |
| ab | AaBb | Aabb | aaBb | aabb |
Lets say each of the genes involves complete dominance and recessiveness. Thus, AA and Aa encode a red phenotype, and aa represents white. Lets also say that BB and Bb are hairy, whereas bb is bald. Lets calculate the genotypes, genotypic ratios, phenotypes, and phenotypic ratios for the above cross:
| GENOTYPIC RATIO | GENOTYPE | PHENOTYPE |
|---|---|---|
| 1 | AABB | red, hairy |
| 2 | AABb | red, hairy |
| 1 | AAbb | red, bald |
| 2 | AaBB | red, hairy |
| 4 | AaBb | red, hairy |
| 2 | Aabb | red, bald |
| 1 | aaBB | white, hairy |
| 2 | aaBb | red, hairy |
| 1 | aabb | white, bald |
Thus, our phenotypic ratio is 9:3:3:1 (9 red and hairy, 3 red and bald, 3 white and hairy, and 1 white and bald). Anytime you perform a dihybrid cross, work with two totally heterozygous individuals, and if the alleles exhibit dominance and recessiveness, then you will end up with these same genotypic and phenotypic ratios. NOTE: One thing that you should notice about the dihybrid crosses is that each pair of alleles of a gene behaves independent of the pair of alleles for the other gene. When Mendel performed his dihybrid crosses, he noticed this as well. Therefore, he formulated his Law of Independent Assortment, which states that the members of one pair of factors (alleles) assort (separate) independently of another pair of factors. Thus, knowing all of the above, lets work some dihydrid cross problems.
PROBLEM 3.
You know that the possession of claws (WW or Ww) is dominant to lack of claws (ww). You also know that the presence of smelly feet (FF or Ff) is dominant to non-smelly feet (ff). You cross a male who is clawed and has smelly feet with a female who is clawed and has non-smelly feet. All 18 offspring produced have smelly feet, and 14 have claws and 4 are un-clawed. What are the genotypes of the parents?
PROBLEM 4.
You have an individual who is totally heterozygous for 2 genes that are not linked (i.e., not on the same chromosome). One gene is for ear size (AA or Aa being big ears whereas aa is for small ears) and the other gene is for buggy eyes (BB and Bb for buggy eyes whereas bb represents normal eyes). If you testcross this individual, what are the resulting genotypes and phenotypes?
PROBLEM 5.
Now then, after you've completed the problem above, lets ignore the Punnett's square and simply look at the 4 types of offspring from the above cross. What if the actual ratios in your testcross were not 1:1:1:1, but were as follows. What would this represent?
| PERCENTAGES | GENOTYPE | PHENOTYPE | 48% | AaBb | Big ears, buggy eyes | 2% | Aabb | Big ears, normal eyes | 2% | aaBb | Small ears, buggy eyes | 48% | aabb | Small ears, normal eyes |
|---|
PROBLEM 6.
The following is a genetic linkage problem involving 4 genes. You want to determine which of the genes are linked, and which occur on separate chromosomes. You cross two true breeding (i.e., remember that this means that they are homozygous) plants that have the following characteristics:
| PLANT 1 | PLANT 2 |
|---|---|
| Red flowers | White flowers |
| Spiny seeds | Smooth seeds |
| Long pollen grains | Short pollen grains |
| Late blooming | Early blooming |
Following the above cross, all of the offspring have red flowers, spiny seeds, long pollen grains, and early blooming (meaning, that these traits are dominant). You then testcross the F1 generation, which you should realize by now are totally heterozygous individuals, and obtain the ratios below. What's going on?
| 49% red-spiny | 25% red-long | 25% red-early | 25% long-early |
|---|---|---|---|
| 1% red-smooth | 25% red-short | 25% red-late | 25% long-late |
| 1% white-spiny | 25% white-long | 25% white-early | 25% short-early |
| 49% white-smooth | 25% white-short | 25% white-late | 25% short-late |
PROBLEM 7.
The following is a genetic linkage problem also involving 4 genes. You want to determine which of the genes are linked, which occur on separate chromosomes, and the distances between the linked genes. You cross 2 true breeding (i.e. homozygous) plants that have the following "unusual" characteristics:
| PLANT 1 | PLANT 2 |
|---|---|
| Red flowers | White flowers |
| Long pollen grains | Short pollen grains |
| Dumb backtalk | Smart backtalk |
| Mean disposition | Nice disposition |
All of the offspring have red flowers, long pollen grains, give smart backtalk, and have a nice disposition (meaning, that these traits are dominant). You then testcross the F1 generation, and obtain the ratios below. How many chromosomes are involved in the linkages, and what are the positions of the linked genes relative to one another?
| 45% red-long | 25% red-dumb | 25% long-dumb | 48% red-mean | 43% long-mean |
|---|---|---|---|---|
| 5% red-short | 25% red-smart | 25% long smart | 2% red-nice | 7% long-nice |
| 5% white-long | 25% white-dumb | 25% short-dumb | 2% white-mean | 7% short-mean |
| 45% white-short | 25% white-smart | 25% short-smart | 48% white-nice | 43% short-nice |
PROBLEM 8.
In the ABO blood system in human beings, alleles A and B are codominant and both are dominant to the O allele. In a paternity dispute, a type AB woman claimed that one of four men was the father of her type A child (the child would be type A with a genotype of either be AA or AO). Which of the following men could be the father of the child on the basis of the evidence given?
PROBLEM 9.
A brown-eyed, long-winged fly is mated to a red-eyed, long-winged fly. The progeny are: 51 long, red ; 53 long, brown ; 18 short, red ; 16 short, brown Using solely the information provided, what are the genotypes of the parents?
PROBLEM 10.
A strange woman has a bizzare condition known as "Cyclops" syndrome, where she has a single eye in the middle of her forehead. The allele for the normal condition (i.e. NO "Cyclops" syndrome) is recessive (cc). Her father is a Cyclops, as well as her mother. Her father's mother was normal. What is the genotype of the strange woman's father?
PROBLEM 11.
In calico cats, there is an X-linked gene with 2 alleles that control fur color. BB is a black female; B'B' is a yellow female; B'B (heterozygous) is a calico female; B' is a yellow male; and B is a black male. You have recently taken over judge Wapner's job on the People's Court and a woman brings in a black female cat that has given birth to 4 calico female kittens and 2 black male kittens. You must decide which of the defendent's male cats is guilty: the black one or the yellow one.
PROBLEM 12.
A common form of red-green color blindness in humans is caused by the presence of an X-linked recessive allele. Given simply that, please answer the following:
PROBLEM 13.
When studying an inheritance phenomenon, a geneticist discovers a phenotypic ratio of 9:6:1 among offspring of a given mating. Give a simple, plausible explanation of the results. How would you test this hypothesis?
PROBLEM 14.
In an epistasis situation, PP or Pp is purple and pp is yellow. CC and Cc encode the ability to produce color whereas cc prevents color production resulting in an albino (i.e., the C allele either allows, or prevents, P from functioning to produce color). Given the following parental matings, provide the ratios of the offspring that are either purple, yellow, or albino. Remember: all offspring must have at least one big C to produce color or they will be albino.
| OFFSPRING RATIOS | ||||
|---|---|---|---|---|
| PARENTAL CROSSES | purple | yellow | albino | PROVIDE EXPLANATIONS FOR EACH OF YOUR ANSWERS |
| PPCC x PPCC |
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| PPCC x ppcc |
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| ppcc x ppCc |
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| Ppcc x PpCc |
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