-- card: 101881 from stack: in -- bmap block id: 105639 -- flags: 0000 -- background id: 2665 -- name: -- part 1 (button) -- low flags: 00 -- high flags: 8003 -- rect: left=205 top=315 right=337 bottom=256 -- title width / last selected line: 0 -- icon id / first selected line: 0 / 0 -- text alignment: 1 -- font id: 0 -- text size: 12 -- style flags: 0 -- line height: 16 -- part name: NEXT ----- HyperTalk script ----- on mouseUp go to next card end mouseUp -- part 2 (button) -- low flags: 00 -- high flags: 8003 -- rect: left=149 top=314 right=337 bottom=199 -- title width / last selected line: 0 -- icon id / first selected line: 0 / 0 -- text alignment: 1 -- font id: 0 -- text size: 12 -- style flags: 0 -- line height: 16 -- part name: PREV. ----- HyperTalk script ----- on mouseUp go back end mouseUp -- part contents for background part 1 ----- text ----- PEDIGREE #4 -- part contents for background part 2 ----- text ----- This pedigree traces the inheritance of two characteristics... color blindness and a deficiency in glucose 6-phosphate dehydrogenase. The dark shading means the presence of the mutant phenotype. Give the genotypes of the numbered individuals. Note: What occurred to produce the genotype in #4? Color blindness is a sex-linked gene. What can you say about the glucose 6-phosphate enzyme deficiency? Think about your answer and then scroll down the window to see if you were correct in your assumptions. Female #1 is heterozygous for the enzyme deficiency while male #3 has the color blind deficiency. Female #2 is heterozygous for both genes which, by their mode of inheritance must lie on diffrent chromosomes. However crossing-over in female #2 produced a chromosome carrying both genes which then appear together in male #4. To cross over in this fashion means that the enzyme deficiency must be on the X chromosome and therefore sex-linked along with the color blind gene. -- part contents for background part 8 ----- text ----- 164