Biology 198
Answers to Mendelian Genetics problems

Updated: 21 August 2000



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?

Bb (heterozygous male) x bb (testcross female)

b b
B Bb Bb
b bb bb

If the brown male had been BB, then all offspring would have been Bb and all brown. However, if the male is Bb as above and you perform a testcross, 50% of all offspring should have the bb genotype and a white phenotype. A testcross to a heterozygous individual should always yield about a 1:1 ratio of the dominant to recessive phenotype. So, both the genotypic and phenotypic ratios here are 50:50.


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.

Cross #1
RR x rr

r r
R Rr Rr
R Rr Rr

You'll note that 100% of all offspring are Rr, which is the genotype. Since the genotype for all offspring are the same, and the Rr genotype encodes a pink color, then 100% of the phenotypes will be pink (NOT red - remember that this is incomplete dominance). Now then, you need to perform a second cross between the offspring. Since all offspring are Rr, then the cross will be Rr x Rr.

Cross #2
Rr x Rr

R r
r Rr rr

You'll note here that all offspring are not pink. Your genotypic ratio is 25% (RR), 50% (Rr), and 25% (rr). The phenotypic ratio is also the same in this case, with 25% red (RR), 50% pink (Rr), and 25% white (rr).


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?

Answer: Start with what you know early in the story: Dad is clawed, so he has at least one big W. You don't know whether his second allele is big W or little w at this point. He also has smelly feet, so again you know he has one big F but you cannot decipher the second allele at this time. Mom is clawed so she has at least one big W, but the other allele remains unknown. She has non-smelly feet, so she has the recessive characters and can only be "ff." So, based on the above, we know this much: Dad is (W ? F ?) and Mom is (W ? f f). OK, lets look at the offspring. All children had smelly feet. If Dad had a hidden little f, then it would match up with Mom's little f's and about about one-half of the children would have ended up with non-smelly feet (ff). That didn't happen, so Dad must be FF (homozygous dominant). Now then, look at any recessive individuals that may be un-clawed. There are four, and all must be ww. Each child got a little w from Dad and the other little w from Mom. So, both parents must be heterozygous (Ww). Note that just like the monohybrid crosses, how important the recessive offspring are in these types of problems. You automatically know that each parent had that hidden recessive allele based solely on the phenotype of the offspring. So, you figured out the problem without any Punnett squares and the parents are as follows: Dad is "WwFF" and Mom is "Wwff"


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?

Answer: Remember that a testcross represents a cross with a totally recessive individual. These types of crosses are useful in weeding out hidden recessive alleles from your unknown. Remember the information on recessives if you don't remember anything else. By knowing the recessive, you automatically know both the phenotype and genotype. In the monohybrid cross, a testcross of a heterozygous individual resulted in a 1:1 ratio. With the dihybrid cross, you should expect a 1:1:1:1 ratio!

AaBb x aabb

ab ab ab ab
AB AaBb AaBb AaBb AaBb
Ab Aabb Aabb Aabb Aabb
aB aaBb aaBb aaBb aaBb
ab aabb aabb aabb aabb

Thus, you get the following...

25% AaBb Big ears, buggy eyes
25% Aabb Big ears, normal eyes
25% aaBb Small ears, buggy eyes
25% aabb Small ears, normal eyes


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?

48% AaBb Big ears, buggy eyes
2% Aabb Big ears, normal eyes
2% aaBb Small ears, buggy eyes
48% aabb Small ears, normal eyes

Answer: Whenever you know that you have a totally heterozygous individual, and you get this type of lopsided percentage during the testcross, you have discovered that the A and B genes are linked (i.e. they occur on the same chromosome). Thus, they are NOT assorting independently as Mendel states in his second law. If they were, you would get the 1:1:1:1 ratios. The genotypes and phenotypes with the small percentages (Aabb and aaBb) represent outcomes that were produced due to "crossing over" (during Meiosis I, some homologous chromosomes broke between the 2 genes and DNA was exchanged). Because the percentage of these oddball recombinants was low, then it is likely that the genes are fairly near one another. If the percentages of these middle two combinations were 10-12% each, then the distance between the genes would be greater. In this case, "A" and "B" are on the same chromosome whereas "a" and "b" occur on the other chromosome (except for the ones that just crossed over).


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:

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

Answer: A little more difficult, but still something you should be able to figure out. Obviously from the above, the red/white flowers and the spiny/smooth seed traits are not assorting independently. If they were, we would see the 1:1:1:1 ratios (25%:25%:25%:25%) represented for the other sets of genes. Therefore, the flower color gene and seed texture are linked. Because of the high percentage of red-spiny and white-smooth, the allele for red flowers and the allele for spiny seeds are on the same homologue (except for 2% of the offspring, which are a result of the crossover). Conversely, the allele for white petal color and the allele for smooth seeds are on the same chromosome (again, except for the 2% of the offspring that are a result of crossing over). Since all of the other crosses are 1:1:1:1, then all other genes are on chromosomes separate from the first 2. Therefore, 3 separate chromosomes are involved.


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:

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

Answer: As you can see from the above, some characteristics between genes do not assort in the 1:1:1:1 fashion. Therefore, they are linked. In the first column, one can see that red/white and long/short are on the same chromosome and are 10 (5 + 5) units apart (see below). Also, red/white and mean/nice in the third column are linked and are 4 (2 + 2) units apart (see below). Since mean/nice and short/long are on the same chromosome as red/white, they too are linked as can be seen in column five and are 14 (7 + 7) units apart (see below). The gene for smart/dumb must exist on a second, separate chromosome by itself.

The arrangement below is the only one possible

CHROMOSOME: ________ mean/nice ________ red/white ___________________ long/short ________

(mean/nice is separated from red/white by 4 linkage units)
(red/white is separated from long/short by 10 linkage units)
(mean/nice is separated from long/short by 14 linkage units)


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?

  1. The Type A father? Answer: In this case, a type A person would have one of the following genotypes: AA or AO. A man with either of these genotypes could be the father as the mother would donate the A allele to the child and either an A allele from the father or an O allele from the father would produce a child with Type A blood.
  2. The Type B father? Answer: In this case a type B father would have either the genotype BB or BO. A man with the genotype BO could be the father as the mother would donate the A allele to the child and an O allele from the father would produce a child with Type A blood.
  3. The Type O father? Answer: In this case a type O person would have the genotype OO. A man with this genotype could be the father as the mother would donate the A allele to the child and an O allele from the father would produce a child with Type A blood.
  4. The Type AB father? Answer: In this case a type AB person would have the genotype AB. A man with this genotype could be the father as the mother would donate the A allele to the child and an A allele from the father would produce a child with Type A (i.e. AA) blood.

NOTE: In this case, none of the men can be excluded from possible paternity. I guess they'll need to do genetic testing.


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?

Answer: In this case, it is easier to look at each locus separately. At the wing locus, we have two long-winged flies crossed to yield 104 long-winged flies and 34 short-winged flies. This is very close to a 3:1 ratio that we would expect from a monohybrid cross. Thus, the parents must be heterozygous (Ll) at the wing-length locus and long wings must be dominant. At the eye color locus, we have a red-eyed fly crossed with a brown-eyed fly to yield 69 brown-eyed flies and 69 red-eyed flies. This is a 1:1 ratio, which is what we would expect from a monohybrid testcross. However, we do not know which is dominant, red eyes or brown eyes. Thus one parent is heterozygous (Rr) and the other parent is homozygous recessive (rr) at the eye color locus. Combining the information from the two loci, possible genotypes for the parents are LlRr for the brown-eyed, long-winged parent and Llrr for the red-eyed, long-winged parent. The other possibility is Llrr for brown-eyed, long-winged and LlRr for red-eyed, long-winged.


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?

Answer: Because the woman's father was a Cyclops, he had to have at least one big C. However, it is unknown if his other allele was big C or little c. But, interestingly enough, her father's mother was normal. Since normal is recessive (cc), then she could only donate a little c to her son. Thus, the bizzare woman's father is heterozygous (Cc).


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.

Answer: Note first that the mother, a black female, only has big Bs to offer. The black male kittens are of no help in the problem as they got their B alleles (each a single B on a single X-chromosome) from their mother. However, the female kittens are calico, and thus are B'B. They couldn't receive the B' allele from their mother since their mother was black; thus, they had a yellow (B') father.


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:

  1. Can two color-blind parents give birth to a normal son or daughter? Answer: No. 100% of the parental alleles are recessive; thus, there are no normal alleles to give to the offspring.
  2. Can two normal parents produce a color-blind daughter? Answer: No. Dad will give all of his daughters a normal allele. Thus, even if Mom has a hidden recessive allele, the worst case senario is that the daughter would be heterozygote.
  3. Can two normal parents produce a color-blind son? Answer: Yes. If Mom has a hidden recessive allele, 50% of the sons will be color-blind. The other 50% will get her normal allele and be normal.


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?

Answer: As 9:6:1 appears to be a variant of the standard 9:3:3:1 ratio you would expect from a dihybrid cross, the simplest explanation is that this result is from a dihybrid cross in which epistasis plays a role. "Epistasis" is when a pair of alleles (i.e. a recessive) pair, cover up the expression of a dominant allele at another locus (i.e., 1 set of alleles is masking another). In this case, you would expect the phenotypes to have the have the genotypes given below.

9/16 A? B?
6/16 A? bb and aaB?
1/16 aabb

However, to better examine this, you would need to perform a series of test crosses to see if the results of your crosses match your predictions. I didn't ask for that in the problem, but the problem below covers this.


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.

PPCC x PPCC 1 0 0 all offspring PPCC and will have at least one big C and one big P
PPCC x ppcc 1 0 0 all offspring PpCc and will have at least one big C and one big P
ppcc x ppCc 0 1 1 one-half ppCc and one-half ppcc
Ppcc x PpCc 3 1 4 6 different possibilities. See below*

*Out of 16 gametes, 2 will be PPCc (purple); 2 will be PPcc (albino); 4 will be PpCc (purple); 4 will be Ppcc (albino); 2 will be ppCc (yellow); and 2 will be ppcc (albino).

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