1. UNIT 4 NS270 NUTRITIONAL ASSESSMENT AND MANAGEMENT Amy Habeck, RD, MS, LDN You’ll need: Calculator Presentation (found in Doc Sharing)

2. Unit 4 Learning Objectives: 1) Practice Calculations 1) Unit 4 project calculations 2) Review- Nutrition and Diet Therapy a) Chapter 7 – Assessment of the Hospitalized Patient 3) Answer your questions

3. Nutrition Screening  Screening of hospitalized patients  Completed within first 24-48 hours  Identifies characteristics known to be associated with nutrition problems Malnutrition Nutritional risk What are some diagnoses that put a patient at increased nutritional risk?  Patients at nutritional risk should have a nutrition assessment

4. Diagnoses Associated with Increased Nutritional Risk  Trauma: fracture, burn, closed head injury, GSW, spinal cord injury, MVA  Dysphagia  Bowel resection  Short bowel syndrome  Small bowel obstruction  Hypoglycemia  FTT  Congenital heart disease  COPD  Anorexia  Cancer  HIV/AIDS  V/D  Anemia  CVA or hemiparesis  GI bleed  Crohn’s disease  Dumping syndrome  Pressure ulcers  Organ transplant  DM  CAD  pancreatitis

5. Case Study-Question 1  How can you estimate her height? Table 7.1 page 219 S=75.00 +(1.91 KH)-(0.17A) Equation for white female >60 y.o. = (substitute numbers and complete calculation) S=stature KH=knee height (cm) A= age 72 year old female History of osteoporosis with compression fracture KH = 16.5 inches

6. Case Study- Question 1 KH=16.5 in x 2.54cm/in=41.91cm A= age=72 S=75.00 +(1.91x KH)-(0.17xA) S=75.00+(1.91x41.91)-(0.17x72) S=75.00+80-12.2 S=142.8 cm S=stature KH=knee height (cm) A= age 72 year old female History of osteoporosis with compression fracture KH = 16.5 inches

7. Case Study – Question 2  Calculate adjusted body weight (ABW)  Equation and example on page 225  Table 7.5 : % of body weight contributed by body parts  =% of body weight = lower leg + foot  = (substitute numbers and complete calculation) Helen Wt = 115 Amputation =Lower leg and foot

8. Case Study – Question 2  Calculate adjusted body weight (ABW)  Equation and example on page 225  Table 7.5 - % of body weight contributed by body parts  ABW= current wt/(100-% of amputation)x 100  = (substitute numbers and complete calculation) Helen Wt = 115 Amputation =Lower leg and foot = 5.3 + 1.8= 7.1%

9. Case Study – Question 2  Calculate adjusted body weight  Equation and example on page 225  Table 7.5 - % of body weight contributed by body parts  Adjusted wt = current wt/(100-% of amputation)x 100  Adjusted wt = 115/(100 - 7.1) x 100  Adjusted wt = 115/(92.9) x 100  Adjusted wt = 123.79 pounds Helen Wt = 115 Amputation =Lower leg and foot = 5.3 + 1.8= 7.1%

10. Calculate BMI based on adjusted body weight- Question 3 P176 Lee and Nieman, classification table 6.6  BMI based on adjusted body weight= =ABW(pounds)/ht(in)/ht(in)x703= = (substitute numbers and complete calculation) Justin Entire right leg amputation Ht: 5’6” = 66” ABW: 178 pounds

11. Calculate BMI based on adjusted body weight- Question 3 P176, classification table 6.6  Ht= 5’6”=66”  BMI based on adjusted body weight= =ABW(pounds)/ht(in)/ht(in)x703= =178/(66) 2 x703=28.7 Justin Entire right leg amputation Ht: 5’6” ABW: 178 pounds

12. Evaluating Desirable Body Weight- Question 4  Hamwi equations Hamwi equation is found on page 170-171 of Lee and Nieman  Men 5’ = 106# + 6# for every additional inch +/- 10%  Women 5’=100# + 5# for every additional inch +/- 10%  Height-weight tables  What are some of the limitations of the height-weight tables?

13. IBW-Number 4  What is her desirable or ideal body weight?  Use the Hamwi equation  = (substitute numbers and complete calculation)  Page 170-171 48 year old patient, Ms. Geneva Female Height: 5’6” Actual body weight: 155#

14. Let’s Practice – Question 4  What is her desirable or ideal body weight?  Women: 5’=100 + (5x6) +/- 10%  IBW=130+/- 10% or 130x.9 to 130x1.1  IBW=117-143# 48 year old patient, Ms. Geneva Female Height: 5’6” Actual body weight: 155#

15. Using Anthropometric Measures to estimate weight-Number 5  Equation: page 224, table 7.3  Female: (MAC x 1.63)+(CCx1.43)-37.46  Male: (MACx2.31)+(CCx1.5)-50.10  Estimated weight =  = (substitute numbers and complete calculation) Marjorie MAC = 30 cm CC = 34 cm

16. Using Anthropometric Measures to estimate weight-Number 5  Equation: page 224, table 7.3  Female: (MAC x 1.63)+(CCx1.43)-37.46  Estimated weight = (MAC x 1.63)+(CCx1.43)-37.46  Estimated weight =(30 x 1.63)+(34x1.43)-37.46=  Estimated weight = 48.9+48.62-37.46=60kg  Estimated weight = 60kgx2.2pounds/kg=132 pounds Marjorie MAC = 30 cm CC = 34 cm

17. Using Anthropometric Measures to estimate weight-Number 6  Known: KH and MAC (both in cm)  Equation: page 225, table 7.4  Convert your answer to pounds  Est. weight = (KHx1.09)+(MAC x 3.14)-83.72  = (substitute numbers and complete calculation) 50 y.o. black male KH = 42 cm MAC = 30 cm

18. Using Anthropometric Measures to estimate weight-Number 6  Known: KH and MAC (both in cm)  Equation: page 225, table 7.4  Convert your answer to pounds  Est. weight = (KHx1.09)+(MAC x 3.14)-83.72  Est. weight = (42x1.09)+(30 x 3.14)-83.72  Est. weight = (45.78)+(94.2)-83.72  Est. weight = 55.96 kg  55.96 kg x 2.2 pound/kg = 123 pounds 50 y.o. black male KH = 42 cm MAC = 30 cm

19. Calculating REE Question 7  Calculate the REE for Ms. Geneva  What results do you get with each of the methods below? Mifflin-St. Jeor Harris-Benedict WHO National Academy of Sciences  How do they compare?  Which method do you like better and why?  Apply an activity factor for average activity to each of your results. List the answer separately. 48 year old patient, Ms. Geneva Female with right below-knee amputation Height: 5’6” Actual body weight: 155#

20. Ms. Geneva- Mifflin-St. Jeor  48 year old female patient, s/p R BKA  Height: 5’6”=66inx2.54cm/in=167.6cm  Actual body weight: 155# x 1kg/2.2#=70.5kg  Female: REE = 10×weight + 6.25×height - 5×age - 161  REE= (10x70.5)+(6.25x167.6)-(5x48)-161  REE= (705)+(1047.5)-(240)-161  REE= 1351.5 kcal/day

21. Ms. Geneva-Harris Benedict  48 year old female patient, s/p R BKA  Height: 5’6”=66inx2.54cm/in=167.6cm  Actual body weight: 155#x1kg/2.2#=70.5kg  p 232, table 7.7  REE=655.1+9.6W+1.9S-4.7A  REE=655.1+(9.6x70.5)+(1.9x167.6)-(4.7x48)  REE=655.1+676.8+318.4-225.6  REE=1425 kcal/day

22. Ms. Geneva- WHO  48 year old female patient, s/p R BKA  Height: 5’6”=66inx2.54cm/in=167.6cm  Actual body weight: 155#x1kg/2.2#=70.5kg  p 232, table 7.7  REE=8.7W+829  REE=8.7x70.5+829  REE=613.4+829  REE=1442kcal/d

23. Ms. Geneva- NAS  48 year old female patient, s/p R BKA  Height: 5’6”=66inx2.54cm/inx1m/100cm=1.68m  Actual body weight: 155#x1kg/2.2#=70.5kg  p 233, table 7.8  REE=247-(2.67xA)+(401.5xht)+(8.6xwt)  REE=246-(2.67x48)+(401.5x1.68)+(8.6x70.5)  REE=245-128.2+674.5+606.3  REE=1398kcal/d

24. Compare  Mifflin-St. Jeor  REE=1351 kcal/d  Harris Benedict  REE=1425 kcal/d  WHO  REE=1442 kcal/d  NAS  REE=1398 kcal/d  How do you think these equations compare?

25. EER-Number 8  P 233-234, table 7.9  EER for males >19 years  EER = (substitute numbers and complete calculation) 29 y.o. male PA=physical activity factor Ht in m= 1.78m wt in kg=90.5 PA=1.11 (low active) age=45

26. EER-Number 8  P 233-234, table 7.9  EER for males >19 years  PA=physical activity factor  Ht in m= 1.78m, wt in kg=90.5, PA=1.11 (low active), age=45  EER=662-9.53(age)+PA x(15.91x wt+539.6 x ht)  EER=662-9.53(45y)+1.11 x(15.91x 90.5kg+539.6 x 1.78m)  EER=662-428.85+1.11 x( 1439.9+ 960.49)  EER=662-428.85+1.11 x( 2400.39)  EER=662-428.85+ 2664.4  EER=2897.55 kcal

27. Calculate TEE for Overweight Adults- Number 9  Complete the calculations for Ms. Geneva using the TEE equation in Table 7.10 on page 235.  48 year old female patient, s/p R BKA  Height: 5’6”=66inx2.54cm/in=167.6cm  Actual body weight: 155#x1kg/2.2#=70.5kg  Use an activity factor for active adults.  = (substitute numbers and complete calculation)

28. Calculate TEE for Overweight Adults- Number 9  48 year old female patient, s/p R BKA  Height: 5’6”=66inx2.54cm/in=167.6cm  Actual body weight: 155#x1#/2.2kg=70.5kg  TEE=448-(7.95xage)+PA x(11.4x Wt + 619x ht)  TEE=448-(7.95x48)+1.27 x(11.4x70.5+619x1.68)  TEE=448-(7.95x48)+1.27 x(803.7+1039.9)  TEE=448-384.6+1.27 x 1843.6  TEE=448-384.6+2341.4  TEE=2405kcal/d

29. Ireton-Jones-number 10  Ireton-Jones recommended for overweight individuals in critical condition  Activity factors (AF): box 7.3, page 233  Injury factors (IF): table 7.11, page 237  Obesity factor: 1=BMI>27, 0=BMI<27  IJEE: 629-11(age)+25(wt)-609(obesity factor)

30. Ireton-Jones Energy Expenditure Number 10  Ireton-Jones Equation  IJEE = 629 - 11 (A) + 25 (W) - 609 (O) A = age in years, W = weight in kg, O= obesity  Obesity : BMI >27 = 1, BMI ≤27 = 0  Kcals = IJEE x AF x IF  Activity factors (AF) - Box 7.3, p. 233  Injury factors (IF) -Table 7.11, p. 237  IJEE = (substitute numbers and complete calculation) 50 year old female, confined to bed after minor surgery Ht: 5’4” Actual body weight: 176# BMI = 30.2

31. Ireton-Jones Energy Expenditure Number 10  Ireton-Jones Equation  IJEE = 629 - 11 (A) + 25 (W) - 609 (O)  IJEE=629 –(11x 50)+(25x80)-609(1)  IJEE=629 – 550+2000-609  IJEE=1470 kcal/day 50 year old female, confined to bed after minor surgery Ht: 5’4” Actual body weight: 176# BMI = 30.2

32. Ireton-Jones Energy Expenditure Number 10  Ireton-Jones Equation  IJEE = 629 - 11 (A) + 25 (W) - 609 (O)  IJEE=629 –(11x 50)+(25x80)-609(1)  IJEE=629 – 550+2000-609  IJEE=1470 kcal/day  Kcals = IJEE x AF x IF  Activity factors (AF) – confined to bed=1.2  Injury factors (IF) –minor surgery=1.0-1.1  Kcals = 1470 x 1.2 x 1.1  Kcals= 1940 kcal/day 50 year old female, confined to bed after minor surgery Ht: 5’4” Actual body weight: 176#

33. Estimating Protein Needs- Number 11  Estimate Ms. Geneva’s protein needs for an individual who has undergone major surgery.  P 239, table 7.13 48 year old patient, Ms. Geneva Female Height: 5’6” Actual body weight: 155#

34. Estimating Protein Needs Number 11  Estimate Ms. Geneva’s protein needs for an individual who has undergone major surgery.  Moderate stress level: 1.2-1.8 gm/kg

35. Estimating Protein Needs Number 11  Estimate Ms. Geneva’s protein needs for an individual who has undergone major surgery.  Moderate stress level: 1.2-1.8 gm/kg  155 # = 70.45 kg  70.45 x 1.2 = 84.54 gm protein/d  70.45 x 1.8 = 126.81 gm protein/d

36. Questions About Assessment of Hospitalized Patients?

37. Farewell  Thank you for your kind attention and participation!  Email any time - ahabeck@kaplan.eduahabeck@kaplan.edu  Follow me on Twitter  @ProfAmyH