Question

5. It is probable that a probability question is in the future From the cross AaBbceDdEell X AABECcDdeer what is the probability of obtaining the genotype AabbCcDDEeff in the progeny? 6. Fill in XY female Cri du Chat Down syndrome XX male Anhydrotic ectodermal dysplasia Patau syndrome Hemophilia Polydactyly Phenylketonuria Achondroplasia Alkaptonuria a) Develop male structure, but experiences infertility b) Develop female structures but may not have ovaries c) Female heterozygotes have patches of skin that lack sweat glands d) Inability to clot blood due to lack of a clotting factor e) Dominant allele results in extra fingers and toes f) Recessive allele prevents metabolism of phenylalanine g) Dominant allele affects cartilage growth needed for bone lengthening h) Intellectual disability, higher risk of Alzheimer disease i) Babies usually do not live beyond a year j) Microcephaly, abnormal larynx anatomy k) Metabolic disease part of phe pathway, black urine 1) 5 disorders that can be detected by karyotype m) X-linked results in males that do not develop sweat glands n) 6 disorders caused by a single gene mutation o) 2 trisomys p) 2 X-linked disorders 7. Ploidy, somy, and more Fill in: RECIPROCAL TRANSLOCATION LOCUS TETRAPLOID MONOPLOID MONOSOMY BARR BODY TRIPLOID TRISOMY a) Position a gene occupies on a chromosome b) X chromosome in mammals seen as heterochromatin c) Aberrant chromosome number in which normally diploid cell has 3 copies of 1 chromosome d) An exchange of non-homologous chromosomes e) Aberrant chromosomal condition in which cell has I copy of a chromosome and 2 copies of all others f) An individual possessing 4 sets of chromosomes 8. Let's not forget sex..... Arrange in the correct chronological order for human male sex structure development: TDF, SRY, testosterone, male embryo, testes, Y chromosome

I need answers for 5 to 8. thank you

Answer 1

5. Probability of obtaining the genotype AabbCcDdEeff= 1/2568. Y chromosome, SRY, TDF, testes, testosterone, Male embryo

5 Parents :- ha B b c c Dd Eeff x AABb C e Del eett. Offspring - Aa bb CeDDEeff. AaXAA AA, AA, AA, a A. Probability of getting Aa= 2 2 2 2 2 Bb x B b BB, 6B, Bb, bb Probability of getting bb = 1 1 1 сех Сc cC, «С, се се Probability of getting Cc = 2 2 = -2 Dd x Dd - DD, Dd , d D, dd Probability of getting DD = 1/144 Eex ee as Eegte, ee, ee Probability of getting Ee= 2 = 1 / 1 ff xffff, ff, ff, ff Probability of getting ff = 2 2 = 1 / 2 So the probability of obtaining the gebotype Aabb Cc D d E eff I x 1 x 4 X 2 4 = 156