Mendelian genetics (2024)

PROBLEM 5.

Now then, after you've completed the problem above, lets ignore the Punnett's square and simplylook at the 4 types of offspring from the above cross. What if the actual ratios in your testcrosswere not 1:1:1:1, but were as follows. What would this represent?

Answer: Whenever you know that you have a totallyheterozygous individual, and you get this type of lopsided percentage during the testcross, youhave 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, youwould get the 1:1:1:1 ratios. The genotypes and phenotypes with the small percentages (Aabband aaBb) represent outcomes that were produced due to "crossing over" (during Meiosis I,some hom*ologous chromosomes broke between the 2 genes and DNA was exchanged). Becausethe percentage of these oddball recombinants was low, then it is likely that the genes are fairlynear one another. If the percentages of these middle two combinations were 10-12% each, thenthe distance between the genes would be greater. In this case, "A" and"B" are on the samechromosome whereas "a" and "b" occur on the other chromosome (except forthe ones that justcrossed over).

PROBLEM 6.

The following is a genetic linkage problem involving 4 genes. You want to determine which ofthe genes are linked, and which occur on separate chromosomes. You cross two true breeding(i.e., remember that this means that they are hom*ozygous) plants that have the followingcharacteristics:

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 F1generation, which you should realize by now are totally heterozygous individuals, and obtain theratios below. What's going on?

Answer: A little more difficult, but still something you should beable to figure out. Obviously from the above, the red/white flowers and the spiny/smooth seedtraits are not assorting independently. If they were, we would see the1:1:1:1 ratios (25%:25%:25%:25%) representedfor the other sets of genes. Therefore, the flower color gene and seed texture are linked. Becauseof the high percentage of red-spiny and white-smooth, the allele for red flowers and the allele forspiny seeds are on the same hom*ologue (except for 2% of the offspring, which are a result of thecrossover). Conversely, the allele for white petal color and the allele for smooth seeds are on thesame 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 fromthe first 2. Therefore, 3 separate chromosomes are involved.

PROBLEM 7.

The following is a genetic linkage problem also involving 4 genes. You want to determine whichof the genes are linked, which occur on separate chromosomes, and the distances between thelinked genes. You cross 2 true breeding (i.e. hom*ozygous) plants that have the following"unusual" characteristics:

All of the offspring have red flowers, long pollen grains, give smart backtalk, and have a nicedisposition (meaning, that these traits are dominant). You then testcross the F1 generation, andobtain the ratios below. How many chromosomes are involved in the linkages, and what are thepositions of the linked genes relative to one another?

Answer: As you can see from the above, some characteristicsbetween genes do not assort in the 1:1:1:1 fashion. Therefore, they are linked. In the firstcolumn, 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 asred/white, they too are linked as can be seen in column five and are 14 (7 + 7) units apart (seebelow). The gene for smart/dumb must exist on a second, separate chromosome by itself.

PROBLEM 8.

In the ABO blood system in human beings, alleles A and B are codominant and both are dominantto the O allele. In a paternity dispute, a type AB woman claimed that one of four men was thefather 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 personwould have one of the following genotypes: AA or AO. A man with either of these genotypescould be the father as the mother would donate the A allele to the child and either an A allele fromthe 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 fatherwould have either the genotype BB or BO. A man with the genotype BO could be the father asthe mother would donate the A allele to the child and an O allele from the father would produce achild with Type A blood.
3. The Type O father? Answer: In this case a type O personwould have the genotype OO. A man with this genotype could be the father as the mother woulddonate the A allele to the child and an O allele from the father would produce a child with Type Ablood.
4. The Type AB father? Answer: In this case a type AB personwould have the genotype AB. A man with this genotype could be the father as the mother woulddonate 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'llneed to do genetic testing.

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 informationprovided, 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 34short-winged flies. This is very close to a 3:1 ratio that we would expect from a monohybridcross. Thus, the parents must be heterozygous (Ll) at the wing-length locus and long wings mustbe dominant. At the eye color locus, we have a red-eyed fly crossed with a brown-eyed fly toyield 69 brown-eyed flies and 69 red-eyed flies. This is a 1:1 ratio, which is what we would expectfrom a monohybrid testcross. However, we do not know which is dominant, red eyes or browneyes. Thus one parent is heterozygous (Rr) and the other parent is hom*ozygous recessive (rr) atthe eye color locus. Combining the information from the two loci, possible genotypes for theparents are LlRr for the brown-eyed, long-winged parent and Llrr for the red-eyed, long-wingedparent. The other possibility is Llrr for brown-eyed, long-winged and LlRr for red-eyed,long-winged.

PROBLEM 10.

A strange woman has a bizzare condition known as "Cyclops" syndrome, where she has a singleeye 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 motherwas normal. What is the genotype of the strange woman's father?

Answer: Because the woman's father was a Cyclops, he had tohave 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 shecould only donate a little c to her son. Thus, the bizzare woman's father is heterozygous (Cc).

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 ablack male. You have recently taken over judge Wapner's job on the People's Court and awoman brings in a black female cat that has given birth to 4 calico female kittens and 2 black malekittens. You must decide which of the defendent's male cats is guilty: the black one or theyellow one.

Answer: Note first that the mother, a black female, only has bigBs to offer. The black male kittens are of no help in the problem as they got their B alleles (eacha 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 wasblack; thus, they had a yellow (B') father.

PROBLEM 12.

A common form of red-green color blindness in humans is caused by the presence of an X-linkedrecessive 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, thereare 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 hiddenrecessive 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.

PROBLEM 13.

When studying an inheritance phenomenon, a geneticist discovers a phenotypic ratio of 9:6:1amongoffspring of a given mating. Give a simple, plausible explanation of the results. How would youtest this hypothesis?

Answer: As 9:6:1 appears to be a variant of the standard 9:3:3:1ratio you would expect from a dihybrid cross, the simplest explanation is that this result is from adihybrid 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 maskinganother). In this case, you would expect the phenotypes to have the have the genotypes givenbelow.

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 theresults of your crosses match your predictions. I didn't ask for that inthe problem, but the problem below covers this.

PROBLEM 14.

In an epistasis situation, PP or Pp is purple and pp is yellow. CC and Cc encode the ability toproduce color whereas cc prevents color production resulting in an albino (i.e., the C allele eitherallows, 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: alloffspring must have at least one big C to produce color or they will be albino.

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

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