Campbell Problem 6

Molecular Genetics Problem 6
6. A space probe discovers a planet inhabited by creatures who reproduce with the same hereditary patterns as those in humans. Three phenotypic characters are height (T = tall, t = dwarf), hearing appendages (A = antennae, a = no antennae), and nose morphology (S = upturned snout, s = downturned snout). Since the creatures were not “intelligent” Earth scientists were able to do some controlled breeding experiments, using various heterozygotes in testcrosses. For a tall heterozygote with antennae, the offspring were tall-antennae, 46; dwarf-antennae 7; dwarf-no antennae 42; tall-no antennae 5. For a heterozygote with antennae and an upturned snout, the offspring were antennae-upturned snout 47; antennae-downturned snout, 2; no antennae-downturned snout, 48: no antennae-upturned snout 3. Calculate the recombination frequencies for both experiments.

Experiment 1 (Frequency/Distance between T and A).

Determine the recombination frequency for the genes controlling Tallness and Antennae:

46 tall-antennae = 46% expected
42 dwarf-no antennae = 42% expected
7 dwarf-antennae = 7% recombinant
5 tall-no antennae = 5% recombinant

Total = 100

Therefore this recombination frequency between genes T and A is 12%

Experiment 2. (Frequency/Distance between A and S)

Determine the recombination frequency for the genes controlling Antennae and Snout:

47 antennae-upturned snout = 47% expected
48 no antennae-downturned snout = 48% expected
2 antennae-downturned snout = 2% recombinant
3 no antennae-upturned snout = 3% recombinant

Total = 100

Therefore this recombination frequency between genes A and S is 5%

BACK 

Campbell Problem 8

Molecular Genetics Problem 8
8. Imagine that a geneticist has identified two disorders that appear to be caused by the same chromosomal defect and are affected by genomic imprinting: blindness and numbness of the limbs. A blind woman (whose mother suffered from numbness) has four children, two of whom, a son and daughter, have inherited the chromosomal defect. If this defect works like Prader-Willi and Angelman syndromes, what disorders do this son and daughter display? What disorders would be seen in their sons and daughters?

In Prader-Willi and Angelman syndromes the type of symptom exhibited in the offspring depends upon which parent contributes the defective chromosome.

In this case children receiving a defective chromosome from the father will suffer from numbness and children receiving a defective chromosome from the mother will be blind.

The pedigree below helps to sort out how the imprinting works.

 

BACK 

Campbell Problem 9

Molecular Genetics Problem 9
9. What pattern of inheritance would lead a geneticist to suspect that an inherited disorder of cell metabolism is due to a defective mitochondrial gene?

 

The disorder would always be inherited from the mother because the mother’s mitochondrial gene is the only one that survives when the zygote is formed. The gamete from the mother contains all the information. The head of the father’s sperm is the only part that survives during fertilization. The tail of the sperm containing the male’s mitochondria (an their genes) is lost when the zygote begins development. Thus it is only from the mother that the disorder can be inherited.

 

Do Brain Cells Run Out of Gas?
Within each cell reside hundreds of tiny gas stations known as mitochondria. These essential organelles generate a large share of the fuel, a molecule called ATP, that cells use to power their biological machinery. There’s a suspicion, admittedly controversial, that problems with these energy-supplying mitochondria contribute to the progression of age-related neurodegenerative illnesses such as Alzheimer’s, Parkinson’s, and Huntington’s diseases, says Douglas C. Wallace of Emory University School of Medicine in Atlanta. In 1993, Wallace and his colleagues reported on comparisons of the mitochondrial DNA of Alzheimer’s patients and that of people without Alzheimer’s, who served as controls. This genetic material, which contains all the instructions necessary for mitochondria to function and replicate, is independent of the DNA found in a cell’s nucleus. Wallace’s group discovered that a particular mutation in mitochondrial DNA showed up in more than 5 percent of Alzheimer’s patients but in less than 1 percent of a random group of people with-out the disease. Studies on animals support the importance of mitochondria in brain disorders. When investigators destroy mitochondria or inhibit the activity of enzymes crucial to mitochondrial function in rats or mice, the rodents develop behavioral or physical attributes of Alzheimer’s, Huntington’s, and Parkinson’s diseases. &emdash; J. Travis

Science News: Aug. 5 • Vol. 148, No. 6

 

BACK 

Campbell Problem 7

Molecular Genetics Problem 7
7. Using the information from problem 6, a further testcross was done using a heterozygote for height and nose morphology. The offspring were tall-upturned nose, 40; dwarf-upturned nose, 9; dwarf-downturned nose, 42; tall-downturned nose, 9. Calculate the recombination frequency from these data; then use your answer from problem 6 to determine the correct sequence of the three linked genes.

Experiment 3. (Frequency/Distance between T and S)

Determine the recombination frequency for the genes controlling Tallness and Snout:

40 tall-upturned snout = 40% expected
42 dwarf-downturned snout = 42% expected
9 dwarf-upturned snout = 9% recombinant
9 tall-downturned snout = 9% recombinant

Total = 100%

Therefore this recombination frequency between genes T and S is 18%

One can determine the relative frequency between genes using the percent frequencies as distances.

The Recombinant relationships from experiments 1-3 are:

Exp. 1 T-A = 12 map units Exp. 2 A-S = 5 map units Exp. 3 T-S = 18 map units

An arrangement that fits the data would be:

BACK

 

Campbell Problem 10

Molecular Genetics Problem 10
10. An aneuploid person is obviously female, but her cells have two Barr bodies. What is the probable complement of sex chromosomes in this individual?

This individual probably is XXX.

The individual is a female. Nondisjunction of sex chromosomes produces a variety of aneuploid conditions in humans. Most of these conditions appear to upset genetic balance less than aneuploid conditions involving autosomes. Extra copies of the X chromosome are deactivated as Barr bodies in the somatic cells. Females with trisomy of the X chromosome (XXX), which occurs about once in approximately 1000 live births, are healthy and cannot be distinguished from XX females except by karyotype.

An Example of nondisjunction:

Klinefelter’s syndrome

49 ,XXXXY

This karyotype shows a variant of Klinefelter’s syndrome.

Individuals with this syndrome are male, typically with the karyotype 47,XXY.

Individuals with Klinefelter’s syndrome exhibit a characteristic phenotype including tall stature, infertility, gynecomastia and hypogonadism.

Aneuploidy above one extra chromosome is usually fatal but because of X-inactivation, which “turns off” all but one X chromosome per cell, the effects of 3 extra chromosomes are reduced.

BACK