Please explain how the answers were obtained as simply as possible. n22 2n-38 ch
ID: 262479 • Letter: P
Question
Please explain how the answers were obtained as simply as possible.
n22 2n-38 chromosomes, mate. 1 9 a. What is the chromosome number in the allediploid? 4 b. What is the chromosome number in the allotetraploid (assume the karyotype is AABB)? 48 c. Whrch of the two plants (allodiploid or allotetraploid) is sterile? 6. A species has 4n -24. How many chromosomes will be found in all gametic cells prcdnacholoing species varats?. a. Hexaploid663 b, nulloso mic for a chromosome; nullo-none c. Haploid n 6 4??4 20p- ? 7. Thinking about the concept of linkage: (write true/false) Trnea. Genes that are located close to one another and present on the same chromosome are linked. Tme b. Linkage can be demonstrated in a dihybrid test cross. Folse c. Linked genes behave as independently assorting genes during meiosis. Irwe S. A baby boy, karyotypically a female (46, XX), is born to a man and a woman. Later d. Linkage can be broken by the mechanism of crossing over. this infant is diagnosed with sex reversal syndrome. Using your knowledge of chromosomal abnormalities and human sex determination, give a reasonably logical explanation that explains how an individual with 46, XX chromosome constitution could exhibit this type of phenotype. a. Give the most likely chromosomal abnormality that occurred in this case. aanslowtann b. Which chromosome(s) did the abnormality affect? chromesorne What specific genes or gene products are likely involved?Explanation / Answer
(5) Let the two species be A and B.
So, For A (2n)= 44 when n= 44/2= 22 and for B (2n)=38 when n= 38/2=19
(2n= diploid number of chromosomes and n= haploid number of chromosome)
a. In allodiploids, the cells contain one copy of chromosomes from species A and one from species B. Therefore, the number of chromosomes in an allodiploid= n of species A + n of species B= 22+19= 41
b.The allotetraploid ( AABB) is created by doubling the chromosome numbers in allodiploid. Therefore, the total number of chromosomes in allotetraploid = number of chromsomes in allodiploid x 2= 41x 2= 82
c. The allodiploid plant would be sterile since it contains an odd number of chromsomes ,i.e, 41 chromosomes. Cells with odd number of chromosomes are sterile due to lack of a homologous pair of chromosome for the extra chromosome during meiotic crossingover.
(6) If the species has 4n= 24, the haploid number of chromosomes would be n= 24/4= 6.
All gametic cells are haloid and therefore conotain "n" number of chromosomes.
a. Hexaploid- Number of chromosomes in a diploid cell of a hexaploid= 6n= 6X 6= 36
Therefore, the hexaploid cell has 2n= 36. Therefore, the gametic cell would have n= 36/2= 18.
b. nullisomics lack a pair of homologous chromosomes. As the given plant has 4 sets of haploid chromosomes, i.e, 4n=24, therefore, each set would lack one chromosomes. THerefore, the nullisome in this case would be represented by 4n-4= 24-4= 20. (diploid number of chromsomes in the nullisomic)
Thus for the nullisomic plant 2n= 20 and the gametic cell would have n number of chromsomes where n= 20/2=10
c. Haploid speciewould have gametes with haploid number of chromosomes, i.e, n=6. (4n=24).
7. Important note : The concept of linkage indicates that genes which are present close to each other on the same chromsome are unable to assort independently and are said to be linked. These genes have a recombination frequency < 50% in comparison to genes that independently assort during recobination and are present on different chromosomes.
a. True.( explained above)
b. True, A dihybrid test cross can be used to demonstrate linkage. The parent with the dominant phenotype is known to be a heterozygote (AaBb) and the tester parent is known to be completely homozygous recessive (aabb). The heterozygous parent with the dominant phenotype, if the genes showed independent assortment, should make four gametes AB, Ab, aB and ab in equal frequencies. Linkage will be detected if there are more testcross progeny who get the parental gametes from the dominant individual than expected by chance alone. ( deviation in frequencies indicates linkage during test cross)
c. False, linked genes do not assort independently during meiosis. Beinf close to one another these genes tend to remaind together during crossing over, resulting in recombination frequencies < 50%.
d. True. Linkage can be broken by crossing over in cases where the chiasma forms between the linked genes resulting in exchange of genetic material during crossing over.
(8) a. The sex- reversal syndrome (SRS) is a result of unequal crossing over between X and Y chromosomes during meiosis in the father. Translocation of SRY from Y chromosome to X chromosome or other autosomes is one of the key factors that induces XX male SRS.
b.
Males typically have one X chromosome and one Y chromosome. Females have two X chromosomes. XX males that are SRY (sex determining region of Y) -positive have two X chromosomes, with one of them containing genetic material from the Y chromosome, making them phenotypically male (or TDF- testis determining factor) but genetically female.
c. The SRY locus is involved which encodes SRY gene involved in initiation of male sex determination in humans.