Yesterday s Picture UNIT 3D

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1 Warm-Up Predict the results of a dihybrid cross between QqHh and QqHh parents if the Q and H genes are very close together on the same chromosome. (LO 3.15) (LO 3.17)

2 Yesterday s Picture

3

4 Mitochondria and chloroplasts have DNA, too, but this non-nuclear DNA is only inherited from the mother Chromosomes, not alleles!

5 Mitochondria and chloroplasts have DNA, too, but this non-nuclear DNA is only inherited from the mother because the sperm ONLY contributes nuclear DNA and nothing else. Example plant cell fertilization Example animal cell fertilization

6 Mitochondria and chloroplasts have DNA, too, but this non-nuclear DNA is only inherited from the mother because the sperm ONLY contributes nuclear DNA and nothing else. A pedigree in which a gene is only passed through the mother represents a mitochondrial gene. female with gene male with gene female without gene male without gene The gene only passes through the mother!

7 Mitochondria and chloroplasts have DNA, too, but this non-nuclear DNA is only inherited from the mother because the sperm ONLY contributes nuclear DNA and nothing else. A pedigree in which a gene is only passed through the mother represents a mitochondrial gene. Non-nuclear DNA is inherited asexually and identically passed down since the dawn of humanity. The gene only passes through the mother!

8 Mitochondria and chloroplasts have DNA, too, but this non-nuclear DNA is only inherited from the mother because the sperm ONLY contributes nuclear DNA and nothing else. A pedigree in which a gene is only passed through the mother represents a mitochondrial gene. Non-nuclear DNA is inherited asexually and identically passed down since the dawn of humanity. And your mitodna looks very similar to the first woman s!

9 CTQ #1 Explain why there is very little genetic variation in mitochondrial DNA within the human population.

10 To determine if an inheritance is Mendelian or non-mendelian: Assume it is Mendelian P p S s P PP Pp 1# 4 S 1# 4 SS Ss 1# 4 1# 4 p Pp pp s Ss ss 1# 1 4 # 4 1# 1 4 # 4

11 To determine if an inheritance is Mendelian or non-mendelian: Assume it is Mendelian Determine the probability of each phenotype (phenotype distribution) P p S s P PP Pp 1# 4 S 1# 4 SS Ss 1# 4 1# 4 p Pp pp s Ss ss 1# 1 4 # 4 1# 1 4 # 4 Phenotype Distribution: purple, smooth: ¾ x ¾ = purple, wrinkled: ¾ x ¼ = white, smooth: ¾ x ¼ = white, wrinkled: ¼ x ¼ =

12 To determine if an inheritance is Mendelian or non-mendelian: Assume it is Mendelian Determine the probability of each phenotype (phenotype distribution) Find the total number of observed progeny born Phenotype Distribution: purple, smooth: ¾ x ¾ = purple, wrinkled: ¾ x ¼ = white, smooth: ¾ x ¼ = white, wrinkled: ¼ x ¼ = The cross resulted in 872 progeny being purple and smooth, 321 being purple and wrinkled, 344 being white and smooth, and 102 being white and wrinkled total

13 To determine if an inheritance is Mendelian or non-mendelian: Assume it is Mendelian Determine the probability of each phenotype (phenotype distribution) Find the total number of observed progeny born Determine the probability of each phenotype as a fraction of the total number Phenotype Distribution: purple, smooth: ¾ x ¾ = purple, wrinkled: ¾ x ¼ = white, smooth: ¾ x ¼ = white, wrinkled: ¼ x ¼ = x 1639 = 922 x 1639 = 307 x 1639 = 307 x 1639 = 102

14 To determine if an inheritance is Mendelian or non-mendelian: Assume it is Mendelian Determine the probability of each phenotype (phenotype distribution) Find the total number of observed progeny born Determine the probability of each phenotype as a fraction of the total number See if the numbers are similar (or, more precisely, do chi-square analysis, which we will discuss tomorrow). The cross resulted in 872 progeny being purple and smooth, 321 being purple and wrinkled, 344 being white and smooth, and 102 being white and wrinkled. Phenotype Distribution: purple, smooth: ¾ x ¾ = purple, wrinkled: ¾ x ¼ = white, smooth: ¾ x ¼ = white, wrinkled: ¼ x ¼ = x 1639 = 922 x 1639 = 307 x 1639 = 307 x 1639 = Meh, pretty close. The closer they are, the less likely the genes are linked, and the further apart they are on the chromosome (or on different chromosomes!)

15 CTQ #2 In fruit flies, b is an autosomal gene encoding body color: flies with the dominant B allele have a tan body, and flies with the recessive b allele have black bodies. The cy gene, which encodes wing structure, is also autosomal. Flies with the dominant Cy allele have curly wings, while flies with two copies of the recessive cy allele have straight wings. Two flies with the genotypes BbCycy and BbCycy have 1,000 progeny: 332 with tan bodies and curly wings; 324 with tan bodies and straight wings; 346 with black bodies and curly wings; and no progeny with black bodies and straight wings. Create a table as below listed observed and expected phenotype distributions, and predict whether this follows the predicted (Mendelian) pattern of inheritance. (LO 3.17) tan body, curly wings tan body, straight wings black body, curly wings black body, straight wings Observed number Expected number

16 Closure Predict the phenotype distributions of a dihybrid cross between a qqhh mother and a QQhh father if the q gene is mitochondrial. (LO 3.17)

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