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Chapter 15: Q15.4-2CC (page 309)

The ABO blood type locus has been mapped on chromosome 9. A father with type AB blood and a mother who has type O blood have a child with trisomy nine and type A blood. Using this information, can you tell in which parent the non-disjunction occurred? Explain your answer. (See Figures 14.11 and 15.13).

Short Answer

Expert verified

The child has a genotype, namely IAIAi or IAii. The sperm produced from the father has a genotype of IAIA. The egg possesses genotype ii from non-disjunction that occurs during meiosis I or meiosis II.

Step by step solution

01

Description of non-disjunction

The process of failure in the separation of homologous chromosomes or sister chromosomes is known as non-disjunction. There is an occurrence of abnormal chromosomes in the offspring. It can result in certain defective or disorder conditions.

02

Description of trisomy 9

Trisomy 9 is the condition in which the occurrence of additional chromosomes in the chromosomal location is 9. It is a rare condition compared to trisomy 21 in Down syndrome. The survival rate of trisomy 9 is lower.

03

Prediction of non-disjunction in child

The child has trisomy 9 conditions that possess genotypes such as IAIAi or IAii.The sperm produced from the father has a non-disjunction that takes place during paternal meiosis II. The non-disjunction is the process that results in the failure of the separation of chromosomes in a proper condition.

The egg having genotype ii possesses a non-disjunction condition during the process of maternal meiosis I or meiosis II.

Hence, there is a possibility that non-disjunction can take place in the father and mother.

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Most popular questions from this chapter

A planet is inhabited by creatures that reproduce with the same hereditary patterns seen in humans. Three phenotypic characters are height (T = tall, t= dwarf), head appendages (A = antennae, a = no antennae), and nose morphology (S = upturned snout, s = downturned snout). Since the creatures are not 鈥渋ntelligent,鈥� Earth scientists are able to do some controlled breeding experiments using various heterozygotes in testcrosses. For tall heterozygotes with antennae, the offspring are tall antennae, 46; dwarf antennae, 7; dwarf no antennae, 42; tall no antennae, 5. For heterozygotes with antennae and an upturned snout, the offspring are 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.

A white-eyed Drosophila is mated with a red-eyed (wild-type) male, the reciprocal cross of the one shown in Fig 15.4. What phenotypes and genotypes do you predict for the offspring from this cross?

Gene A, B, and C are located on the same chromosome. Test crosses show that the recombination frequency between A and B is 28% and that between A and C is 12%. Can you determine the linear order of these genes?

A wild-type fly (heterozygous for gray body and normal wings) is mated with a black fly with vestigial wings. The offspring have the following phenotypic distribution: wild type, 778; black vestigial; 785; black normal, 158; gray vestigial, 162. What is the recombination frequency between these genes for the body color and wing size? Is this consistent with the results of the experiment in Figure 15.9?

Propose a possible reason that the first naturally occurring mutant fruit fly Morgan saw involved a gene on a sex chromosome and was found a male.
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